bn ghi lại đè đi ạ

đề cứ sai sai

Ảo diệu như hay.

ĐKXĐ: \(x\le2\)

\(PT\Leftrightarrow\left(2\sqrt{2-x}+3\right)\left(1-\sqrt{2-x}\right)\left(3-4x-2\sqrt{2-x}\right)=0\)

...

a) \(\Leftrightarrow x^4-4x-1=0\)

\(\Leftrightarrow x^4+2x^2+1-2x^2-4x-2=0\)

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

\(\Leftrightarrow\left(x^2+1\right)^2=2\left(x+1\right)^2\)

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

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

Tự giải pt bậc 2 nhak :))))

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

Lời giải:

$4x-6=2x+4$

$\Leftrightarrow (4x-6)-(2x+4)=0$

$\Leftrightarrow 2x-10=0$

$\Leftrightarrow 2x=10$

$\Leftrightarrow x=5$

là như nào ạ em chưa hiểu lắm

a/

\(\left(x-1\right)^2-\left(x+1\right)^2=2x-6\\ x^2-2x+1-\left(x^2+2x+1\right)=2x-6\\ \)

\(\Leftrightarrow x^2-2x+1-x^2-2x-1-2x+6=0\)

\(\Leftrightarrow6-6x=0\)

=> x=1

Làm có tâm ghê :)

+)x=0 khong phai la nghiem cua phuong trinh

+)chia ca 2 ve cho \(x^2\ne\) 0 ta co:

\(x^2-5x+8-\frac{5}{x}+\frac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)+8=0\)         (1)

Dat \(x+\frac{1}{x}=a\)   \(\left(\left|a\right|\ge2\right)\)

\(\Rightarrow\)\(x^2+\frac{1}{x^2}=a^2-2\)

(1)\(\Leftrightarrow\)\(\left(a^2-2\right)-5a+8=0\)

den day ban tu giai tiep nhe