(7x-2-50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=7,\frac{1}{7}\). (7x-2-50x+7=(7x-1)(x-7)), so \(x=\frac{1}{7}\) and (7). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(7x-2-50x+7=(7x-1)(x-7)), इसलिए \(x=\frac{1}{7}\) और (7) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
Multiplying both sides by (7x) gives \(7+7x^2=50x\), that is \(7x^2-50x+7=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(7x^2-50x+7=0\). Multiplying both sides by (7x) gives \(7+7x^2=50x\), that is \(7x^2-50x+7=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (7x) से गुणा करने पर \(7+7x^2=50x\), यानी \(7x^2-50x+7=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
(6x-2-37x+6=(6x-1)(x-6)), so \(x=\frac{1}{6}\) and (6). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=6,\frac{1}{6}\). (6x-2-37x+6=(6x-1)(x-6)), so \(x=\frac{1}{6}\) and (6). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(6x-2-37x+6=(6x-1)(x-6)), इसलिए \(x=\frac{1}{6}\) और (6) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
Multiplying both sides by (6x) gives \(6+6x^2=37x\), that is \(6x^2-37x+6=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(6x^2-37x+6=0\). Multiplying both sides by (6x) gives \(6+6x^2=37x\), that is \(6x^2-37x+6=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (6x) से गुणा करने पर \(6+6x^2=37x\), यानी \(6x^2-37x+6=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
(5x-2-26x+5=(5x-1)(x-5)), so \(x=\frac{1}{5}\) and (5). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=5,\frac{1}{5}\). (5x-2-26x+5=(5x-1)(x-5)), so \(x=\frac{1}{5}\) and (5). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(5x-2-26x+5=(5x-1)(x-5)), इसलिए \(x=\frac{1}{5}\) और (5) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
Multiplying both sides by (5x) gives \(5+5x^2=26x\), that is \(5x^2-26x+5=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(5x^2-26x+5=0\). Multiplying both sides by (5x) gives \(5+5x^2=26x\), that is \(5x^2-26x+5=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (5x) से गुणा करने पर \(5+5x^2=26x\), यानी \(5x^2-26x+5=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
(4x-2-17x+4=(4x-1)(x-4)), so \(x=\frac{1}{4}\) and (4). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=4,\frac{1}{4}\). (4x-2-17x+4=(4x-1)(x-4)), so \(x=\frac{1}{4}\) and (4). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(4x-2-17x+4=(4x-1)(x-4)), इसलिए \(x=\frac{1}{4}\) और (4) हैं। परीक्षा में मिले हल मूल समीकरण में मान्य हैं या नहीं जांचें।
Multiplying both sides by (4x) gives \(4+4x^2=17x\), that is \(4x^2-17x+4=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(4x^2-17x+4=0\). Multiplying both sides by (4x) gives \(4+4x^2=17x\), that is \(4x^2-17x+4=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (4x) से गुणा करने पर \(4+4x^2=17x\), यानी \(4x^2-17x+4=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
(3x-2-10x+3=(3x-1)(x-3)), so \(x=\frac{1}{3}\) and (3). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=3,\frac{1}{3}\). (3x-2-10x+3=(3x-1)(x-3)), so \(x=\frac{1}{3}\) and (3). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(3x-2-10x+3=(3x-1)(x-3)), इसलिए \(x=\frac{1}{3}\) और (3) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
Multiplying both sides by (3x) gives \(3+3x^2=10x\), that is \(3x^2-10x+3=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-10x+3=0\). Multiplying both sides by (3x) gives \(3+3x^2=10x\), that is \(3x^2-10x+3=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (3x) से गुणा करने पर \(3+3x^2=10x\), यानी \(3x^2-10x+3=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
(2x-2-5x+2=(2x-1)(x-2)), so \(x=\frac{1}{2}\) and (2). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=2,\frac{1}{2}\). (2x-2-5x+2=(2x-1)(x-2)), so \(x=\frac{1}{2}\) and (2). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(2x-2-5x+2=(2x-1)(x-2)), इसलिए \(x=\frac{1}{2}\) और (2) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
Multiplying both sides by (2x) gives \(2+2x^2=5x\), that is \(2x^2-5x+2=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-5x+2=0\). Multiplying both sides by (2x) gives \(2+2x^2=5x\), that is \(2x^2-5x+2=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (2x) से गुणा करने पर \(2+2x^2=5x\), यानी \(2x^2-5x+2=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
