a/ Áp dụng tính chất phân phối ta được:

\(\left(x+1\right)\left(x+2\right)\)

\(=x^2+x+2x+2\)

\(=x^2+2x+1^2+x+1\)

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

Mà \(x< \left(x+1\right)^2\)

\(\Rightarrow\left(x+1\right)^2+x+1>0\)

=> Biểu thức trên lớn hơn 0

=> Không có kết quả (Sai đề)

b/ Áp dụng tính chất phân phối ta được:

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

\(=x^2-2x+\frac{2}{3}x-\frac{4}{3}\)

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

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

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

Mà \(\left(x-1\right)^2\ge0\)

=> Để thỏa mãn đề bài cần \(\frac{1}{3}\left(2x-1\right)>0\)

=> \(2x>1\Rightarrow x>\frac{1}{2}\)

a ) \(\left(x+1\right).\left(x+2\right)< 0\)

\(=x.\left(x+2\right)+1.\left(x+2\right)< 0\)

\(=x.\left(x-2\right)+\left(x+2\right)< 0\)

\(\Rightarrow x\in Z\)

\(\Rightarrow x>2\)

b ) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\)

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

\(=x+\frac{2}{3}\) = Số nguyên

Nên x thuộc phân số

a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)

hôm sau mik giải tip cho

a) \(\left(2\sqrt{x}-3\right)\left(2+\sqrt{x}\right)+6=0\left(ĐK:x\ge0\right)\)

\(\Leftrightarrow4\sqrt{x}+2x-6-3\sqrt{x}+6=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=0\\2\sqrt{x}+1=0\left(loại\right)\end{array}\right.\)\(\Leftrightarrow x=0\)

b)\(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(ĐK:x\ge3\right)\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\left(tm\right)\\x=6\left(tm\right)\end{array}\right.\)

     \(x^4-2x^3-2x^2+3x+2=0\)

\(\Leftrightarrow x^4-2x^3-2x^2+4x-x+2=0\)

\(\Leftrightarrow\left(x^4-2x^3\right)-\left(2x^2-4x\right)-\left(x-2\right)=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(x^3-2x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-x-x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x^3-x\right)-\left(x+1\right)\right]=0\)

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

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

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

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

Đến đây ez r

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy x=0 hoặc x=4 là giá trị cần tìm

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

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

vậy phương trình có tập nghiệm là S={0;4}

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Rightarrow}}\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)