a)\(\left|x+\frac{1}{5}\right|-4=-2\)

\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)

\(\Rightarrow x+\frac{1}{5}=2\) hoặc \(-2\)

Xét \(x+\frac{1}{5}=2\Leftrightarrow x=\frac{9}{5}\)

Xét \(x+\frac{1}{5}=-2\Leftrightarrow x=-\frac{11}{5}\)

phần a dấu + fai là dấu =

e)

=> (x-2) . (x+7) = ( x-1 ) . ( x +4)

=> x² +7x - 2x -14 = x² - x + 4x - 4

x² + 5x - 14 = x² + 3x - 4

=> 5x - 14  = 3x - 4

=> 5x  - 3x = 14-4

=> 2x         = 10 => x = 10 : 2 => x = 5

c)

=>( x-1) . 7 = ( x + 5 ) . 6

=> 7x - 7 = 6x + 30

=> 7x - 6x=  30 + 7

=> x         = 37

a,x=\(\frac{5}{2}\)

b,x=\(\frac{13}{176}\)

c,x=37

d, x=\(\frac{12}{5}\)

e, x=5

bn đăng từng câu 1 thôi nhe

1) \(\left(x-2\right)\left(\frac{x+1}{3}-x+1\right)=0\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{3}-x^2+x-\frac{2\left(x+1\right)}{3}+2x-2=0\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{3}-x^2+3x-\frac{2\left(x+1\right)}{3}-2=0\)

\(\Leftrightarrow x\left(x+1\right)-3x^2+9x-2\left(x+1\right)-6=0\)

\(\Leftrightarrow x^2+x-3x^2+9x-2x-2-6=0\)

\(\Leftrightarrow-2x^2+8x-8=0\)

\(\Leftrightarrow-2\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow-2.\left(x^2-2.x.2+2^2\right)=0\)

\(\Leftrightarrow-2\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy nghiệm của phương trình là: {2}

2) \(\left(3x+4x\right)\left(\frac{x}{2}-x-\frac{3x}{5}+1\right)=0\)

\(\Leftrightarrow7x\left(\frac{x}{2}-x-\frac{3x}{5}+1\right)=0\)

\(\Leftrightarrow7x\left(-\frac{11x}{10}+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7x=0\\-\frac{11x}{10}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{11}{10}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{10}{11}\end{cases}}\)

Vậy: nghiệm của phương trình là: \(\left\{0;\frac{10}{11}\right\}\)

3) \(\left|x-1\right|=x^2-x\)

\(\Leftrightarrow x-1=x^2-x\)

\(\Leftrightarrow1=x^2-x-x\)

\(\Leftrightarrow1=x^2\)

\(\Leftrightarrow x^2=1\)

\(\Rightarrow x=\pm1\)

Vậy nghiệm phương trình là: {1; -1}

4) \(\left|x^2-3x+1\right|=2x-3\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-3x+1=2x-3\\x^2-3x+1=-\left(2x-3\right)\end{cases}}\)

Xét  trường hợp này rồi làm tiếp, dễ rồi :))

1)\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)

\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow\left(x+1\right)=0\).Do \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\ne0\)

\(\Leftrightarrow x=-1\)

\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\\ \Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\\ \Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

Gọi \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)\)là A

Vì A ≠0

⇒\(x+1=0\Rightarrow x=-1\)

Vậy \(x=-1\)

\(1a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

\(\Leftrightarrow36x=-3\)

\(x=-\frac{1}{12}\)

Vậy ................

\(b,\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

\(\Leftrightarrow\frac{5\left(7x-1\right)}{30}+\frac{30.2x}{30}=\frac{6\left(16-x\right)}{30}\)

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

Vậy ....................

Bài 1:

c) \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

\(\Leftrightarrow\frac{8.\left(x-2\right)^2}{8.3}-\frac{3.\left(2x-3\right).\left(2x+3\right)}{3.8}+\frac{4.\left(x-4\right)^2}{4.6}=0\)

\(\Leftrightarrow\frac{8.\left(x^2-4x+4\right)}{24}-\frac{3.\left(4x^2-9\right)}{24}+\frac{4.\left(x^2-8x+16\right)}{24}=0\)

\(\Rightarrow8.\left(x^2-4x+4\right)-3.\left(4x^2-9\right)+4.\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow8x^2-32x+32-\left(12x^2-27\right)+4x^2-32x+64=0\)

\(\Leftrightarrow8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)

\(\Leftrightarrow123-64x=0\)

\(\Leftrightarrow64x=123-0\)

\(\Leftrightarrow64x=123\)

\(\Leftrightarrow x=123:64\)

\(\Leftrightarrow x=\frac{123}{64}.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{123}{64}\right\}.\)

Chúc bạn học tốt!

Bài 1:

a) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2-2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)

\(\Leftrightarrow36x+3=0\)

\(\Leftrightarrow x=12\)

Vậy phương trình có nghiệm là x = 12

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

=> x \(\in\) {-1;0;1;2;3;4;5;6}

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)

\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

\(\Leftrightarrow-1\le x< 7\)

Mà x nguyên

=>x={-1;0;1;2;3;4;5;6}