Câu 1 :

Xét điều kiện:\(\hept{\begin{cases}x\ge5\\x\le1\end{cases}}\)(Vô lý) 

Vậy pt vô nghiệm

Câu 2 : 

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

Vậy x=-1

Câu 3 : 

\(\sqrt{3x^2-4x+3}=1-2x\)\(\Leftrightarrow3x^2-4x+3=1+4x^2-4x\)

\(\Leftrightarrow x^2=2\Leftrightarrow x=\sqrt{2}\)

Câu 4 : 

\(4\sqrt{x+1}-3\sqrt{x+1}=4\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x=15\)

\(\sqrt{-4x^2+25}=x\)

\(\Leftrightarrow\sqrt{-4x^2+25}^2=x^2\)

\(\Leftrightarrow-4x^2+25=x^2\)

\(\Leftrightarrow-5x^2=-25\)

\(\Leftrightarrow x^2=5\Leftrightarrow x=\pm\sqrt{5}\)

ban Marakai ko co dieu kien ma da binh phuong 2 ve len roi a

\(\sqrt{\frac{-6}{1+x}}=5\)

\(\Leftrightarrow\sqrt{\frac{-6}{1+x}}^2=5^2\)

\(\Leftrightarrow\frac{-6}{1+x}=25\)

\(\Leftrightarrow x+1=\frac{-6}{25}\)

\(\Leftrightarrow x=\frac{-6}{25}-1=\frac{-31}{25}\)

\(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}=2\)

\(\Leftrightarrow\sqrt{x-49}=2\)

\(\Leftrightarrow x-49=4\Leftrightarrow x=53\)

\(a^2+b^2+c^2=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)

\(\Leftrightarrow a^2+b^2+c^2=a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\)

\(\Leftrightarrow a^2+b^2+c^2-2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=4\left(ab+bc+ac\right)\)

\(\Leftrightarrow\left(a+b+c\right)^2=2^2\left(ab+bc+ac\right)\)

\(\Leftrightarrow\left(\frac{a+b+c}{2}\right)^2=ab+bc+ac\)

Suy ra ab+bc+ca là số chính phương

a) Hai đường thẳng y=(3m + 2)x +m-1 và y=(3-m)x-m+2 cắt nhau \(\Leftrightarrow3m+2\ne3-m\)

\(\Leftrightarrow4m\ne1\Leftrightarrow m\ne\frac{1}{4}\)

b) Hai đường thẳng y=(3m + 2)x +m-1 và y=(3-m)x-m+2 song song với nhau \(\Leftrightarrow3m+2=3-m\)

\(\Leftrightarrow4m=1\Leftrightarrow m=\frac{1}{4}\)

P/s: E ms lớp 6, sai thông cảm

\(Q=\frac{2x+2\sqrt{x}+2}{-\sqrt{x}}+\sqrt{x}\)

\(Q=-2\sqrt{x}-2-\frac{2}{\sqrt{x}}+\sqrt{x}\)

\(Q=-\sqrt{x}-\frac{2}{\sqrt{x}}-2\)

\(\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{2}\Rightarrow-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)\le-2\sqrt{2}\)

\(\Rightarrow Q\le-2\sqrt{2}-2\)

\("="\Leftrightarrow x=\sqrt{2}\)