Đề bài là j vậy bạn ???

a) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2x+3\\x+\frac{1}{2}=-\left(2x+3\right)\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x-x=\frac{1}{2}-3\\x+\frac{1}{2}=-2x-3\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x+2x=-3-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\3x=\frac{-7}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x=\frac{-7}{6}\end{array}\right.\)

Vậy \(x\in\left\{\frac{-5}{2};\frac{-7}{6}\right\}\)

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

\(Ta\) \(có\): \(x+\frac{1}{2}=2x+3\)

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

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

\(\Rightarrow\frac{1}{2}=x+3\)

\(\Rightarrow x=\frac{1}{2}-3\)

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

Vậy \(x=-\frac{5}{2}\)

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

\(Ta\) \(có\)

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

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

\(3x+2=4x\)

\(3x+2=3x+x\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

b) Vì \(\left|x+\frac{1}{5}\right|\ge0;\left|x+\frac{2}{5}\right|\ge0;\left|x+1\frac{2}{5}\right|\ge0\forall x\)

\(\Rightarrow4x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\) ta có:

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

\(\Rightarrow3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)

\(\Rightarrow2=4x-3x=x\)

Vậy x = 2

Thôi ạ, câu này ez bỏ sừ

\(A^2-3A+2=\left(A-1\right)\left(A-2\right)=\left(\frac{a}{b}+\frac{b}{a}-1\right)\left(\frac{a}{b}+\frac{b}{a}-2\right)\)

\(=\frac{\left(a^2-ab+b^2\right)}{ab}.\frac{\left(a^2-2ab+b^2\right)}{ab}=\frac{\left(a^2-ab+b^2\right)\left(a-b\right)^2}{2a^2b^2}\)

\(=\frac{\left[\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}\right]\left(a-b\right)^2}{2a^2b^2}\ge0\) \(\forall a;b\)

Dấu "=" xảy ra khi và chỉ khi \(a=b\ne0\)

\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)

b: Ta có: \(3^x+3^{x+2}=20\)

\(\Leftrightarrow3^x\cdot10=20\)

\(\Leftrightarrow3^x=2\left(loại\right)\)

Với a; b > 0

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}=\frac{4}{3}\)

\(ab\le\frac{\left(a+b\right)^2}{4}=\frac{9}{4}\)=> \(\frac{1}{ab}\ge\frac{4}{9}\)

Khi đó: \(S=\left(1+\frac{2}{a}\right)\left(1+\frac{2}{b}\right)=1+2\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{4}{ab}\ge1+2.\frac{4}{3}+4.\frac{4}{9}=\frac{49}{9}\)

Dấu "=" xảy ra <=> a = b = 3/2 

vậy min S = 49/9

\(\left(a+b\right)^2=\left(a-b\right)^2+4ab=2^2+4.3=4+12=16\)

\(\left(a+b\right)^2=\left(a-b\right)^2+4ab==2^2+4\cdot3=16\)

\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)

\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)

__

`(x+2)(5-3x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)

__

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

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

1) a.Từ\(\frac{x}{y}=\frac{11}{7}\Rightarrow\frac{x}{11}=\frac{y}{7}\) 

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

\(\frac{x}{11}=\frac{y}{7}=\frac{x-y}{11-7}=\frac{12}{4}=3\)

\(\Rightarrow x=3.11=33;y=3.7=21\)

b) \(\sqrt{2x-3}=5\)

           \(2x-3=25\)

                    \(2x=28\)

                      \(x=14\)

2) a) \(\frac{3}{2}-\frac{5}{6}:\left(\frac{1}{2}\right)^2+\sqrt{4}=\frac{3}{2}-\frac{5}{6}:\frac{1}{4}+2\)

                                                          \(=\frac{3}{2}-\frac{10}{3}+2\)

                                                         \(=\frac{1}{6}\)

_Học tốt nha_

1. a, \(\frac{x}{y}=\frac{11}{7}\)và x-y=12

\(\Rightarrow\frac{x}{11}=\frac{y}{7}\)và x-y=12

Áp dung tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{11}=\frac{y}{7}=\frac{x-y}{11-7}=\frac{12}{4}=3\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{11}=3\\\frac{y}{7}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=33\\y=21\end{cases}}\)

Vậy

b,\(\sqrt{2x-3}\)=5

\(\Rightarrow2x-3=25\)

\(\Rightarrow2x=28\)

\(\Rightarrow x=14\)

c,\(\frac{3}{2}-\frac{5}{6}:\left(\frac{1}{2}\right)^2+\sqrt{4}\)

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

\(=\frac{3}{2}-\frac{10}{3}+2\)

\(=\frac{9}{6}-\frac{20}{6}+2\)

\(=\frac{-11}{6}+2\)

\(=\frac{1}{6}\)