Tôi nghĩ là như này :)) Sai thì chịu nhá :((

Ta có pt : \(\left|x+1\right|+3\left|x-1\right|=x+2+\left|x\right|+2\left|x-2\right|\) (1)

Ta thấy VT pt (1) là : \(\left|x+1\right|+3\left|x-1\right|\ge0\forall x\)

Nên VP pt (1) cũng phải lớn hơn bằng 0

Có nghĩa là \(x+2\ge0\) \(\Leftrightarrow x\ge-2\)

Khi đó : \(\left\{{}\begin{matrix}\left|x+1\right|=-\left(x+1\right)\\3\left|x-1\right|=3\left(1-x\right)\\\left|x\right|=-x\\2\left|x-2\right|=2\left(2-x\right)\end{matrix}\right.\)

Vậy pt (1) \(\Leftrightarrow-x-1+3-3x=x+2-x+4-2x\)

\(\Leftrightarrow2x=-4\Leftrightarrow x=-2\) ( thỏa mãn )

Vậy \(x=-2\) thỏa mãn pt.

-1 0 1 2

1) x=-2/3>-1( loại)

2)

Thay x = -2 vào phương trình, ta có:

\(4.\left(-2\right)^2-25+q^2+4q.\left(-2\right)=0\)

\(\Leftrightarrow q^2-8q-9=0\Leftrightarrow\left(q-9\right)\left(q+1\right)=0\Leftrightarrow\orbr{\begin{cases}q=-9\\q=1\end{cases}}\)

\(M=x^2-2xy+4y^2+12xy+22\)

\(M=\left(x^2-2xy+y^2\right)+\left(3y^2+12y+12\right)+10\)

\(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=-2\) 

( Chỗ \(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\) bạn phân tích từng cái đã nhá, mình làm tắt ) 

Ta có: \(x^4-30x^2+31x-30=0\) \(\Rightarrow x^4+x-30x^2+30x-30=0\)

\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)

Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

\(\Rightarrow x^2+x-30=0\Rightarrow x^2-5x+6x-30=0\)

\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)

Vậy x=5 hoặc x = -6

a ) Do \(x^2+1>0;\left(x^2+1\right)\left(x-3\right)< 0\Rightarrow x-3< 0\Rightarrow x< 3\)

b ) Do \(x^2+1>0\Rightarrow-\left(x^2+1\right)< 0\)

Mà \(\left(-x^2-1\right)\left(x-1\right)\ge0\Rightarrow x-1\le0\Rightarrow x\le1\)

\(\Leftrightarrow x^4-5x^3+5x^3-25x^3-5x^3+25x+6x-30=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x^3+6x^2-x^2-6x+x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)

hay \(x\in\left\{5;-6\right\}\)

\(\sqrt{x-2\sqrt{x-3}-2}=1\)

=> \(x-2\sqrt{x-3}=1^2=1\)

=> \(-2\sqrt{x-3}=1-x+2\)

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

=> \(\left(-2\sqrt{x-3}\right)^2=\left(3-x\right)^2\)

=> \(4\left(x-3\right)=9-6x+x^2\)

=> \(4x-12=9-6x+x^2\)

=> \(4x-12-9+6x-x^2=0\)

=> \(10x-21-x^2=0\)

Mình xin hết ( biết có vậy )

\(\sqrt{x-2\sqrt{x-3}+2}=1\)

\(\Leftrightarrow\sqrt{x-3-2\sqrt{x-3}+1}=1\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-3}-1\right)^2}=1\)

\(\Leftrightarrow\left|\sqrt{x-3}-1\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}-1=1\\\sqrt{x-3}-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

Vậy....

Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\)

\(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)

\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)

Đặt \(x+\frac{1}{x}=a\) (\(\left|a\right|\ge2\)) \(\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(6\left(a^2-2\right)+7a-36=0\)

\(\Leftrightarrow6a^2+7a-48=0\)

Nghiệm xấu