1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~

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

\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)

Mấy bài kia sao cái phương trình dài thê,s giải sao nổi

2a^4=(1-a)^2=a^2-2a+1

\(A=\frac{2a-3}{\sqrt{2\left(a^2-4a+4\right)}+2a^2}=\frac{2a-3}{\sqrt{2}!\left(a-2\right)!+2a^2}\)a> 2 không thể là nghiệm=> a<2

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

\(A=\frac{2a-3}{\sqrt{2}\left(3\right)}\)

bạn giải thích rõ hơn được không ?

a là nghiệm =>\(\sqrt{2}a^2+a-1=0\Rightarrow\sqrt{2}a^2=1-a\\\)\(2a^4=\left(1-a\right)^2=1^2-2a+a^2\)

Thay 2a^4=...vào ==> 

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

Ta có : \(ax^2+3\left(a+1\right)x+2a+4=0\left(a=a;b=3a+3;c=2a+4\right)\)

Theo hệ thức Vi et ta có : \(x_1+x_2=\frac{-3a-3}{a};x_1x_1=\frac{2a+4}{a}\)

Theo bài ra ta có : \(x_1^2+x_2^2=4\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=4\) Thay vào ta đc : 

\(\Leftrightarrow\left(\frac{-3a-3}{a}\right)^2-2\left(\frac{2a+4}{a}\right)=4\)

\(\Leftrightarrow\frac{9\left(a+1\right)^2}{a^2}-\frac{4a+8}{a}=4\Leftrightarrow\frac{9\left(a+1\right)^2}{a^2}-\frac{4a^2+8a}{a^2}=\frac{4a^2}{a^2}\)

Khử mẫu ta đc : \(9\left(a+1\right)^2-4a^2+8a=4a^2\)

\(\Leftrightarrow9\left(a^2+2a+1\right)-4a^2+8a=4a^2\)

\(\Leftrightarrow9a^2+18a+9-4a^2+8a-4a^2=0\)

\(\Leftrightarrow a^2+27a+9=0\)Ta có : \(\Delta=27^2-4.9=729-36=613>0\)

Nên phương trình có 2 nghiệm phân biệt 

\(x_1=\frac{-27-\sqrt{613}}{2};x_2=\frac{-27+\sqrt{613}}{2}\)

Thay a vào m là xog, tk:

a) \(2\left(x-1\right)-a\left(x-1\right)=2a+3\)

\(\Leftrightarrow2a-2-ax+a=2a+3\)

\(\Leftrightarrow-2-ax+a=3\)

\(\Leftrightarrow-a\left(x-1\right)=5\)

\(\Leftrightarrow\left(x-1\right)=\frac{-5}{a}\Leftrightarrow x=\frac{a-5}{a}\)

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

\(\Leftrightarrow\frac{12x+12+8x+16+6x+18}{24}=3\)

\(\Leftrightarrow12x+12+8x+16+6x+18=72\)

\(\Leftrightarrow26x+46=72\)

\(\Leftrightarrow26x=26\Leftrightarrow x=1\)

c) \(ĐKXĐ:x\ne2;x\ne5\)

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

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

\(\Rightarrow3x\left(x-5\right)-x\left(x-2\right)+3x=0\)

\(\Rightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x^2-10x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

Bài 1: 

c) |2x - 1| = x + 2

<=> 2x - 1 = +(x + 2) hoặc -(x + 2)

* 2x - 1 = x + 2      

<=> 2x - x = 2 + 1

<=> x = 3

* 2x - 1 = -(x + 2)

<=> 2x - 1 = x - 2

<=> 2x - x = -2 + 1

<=> x = -1

Vậy.....

a) với a = -2 ta được phương trình:

3.[(-2) - 2].x + 2.(-2).(x - 1) = 4.(-2) + 3

<=> 3.(-4x) - 4.(x - 1) = (-8) + 3

<=> -12x - 4(x - 1) = -5

<=> -12x - 4x + 4 = -5

<=> -16x + 4 = -5

<=> -16x = -5 - 4

<=> -16x = -9

<=> x = 9/16

b) để x = 1, ta có:

3.(a - 2).1 + 2a(1 - 1) = 4a + 3

<=> 3(a - 2) + 0 = 4a + 3

<=> 3a - 6 = 4a + 3

<=> 3a - 6 - 4a = 3

<=> -a - 6 = 3

<=> -a = 3 + 6

<=> a = -9