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khongg hiểu à ? 

\(A=\frac{1-x^2}{x}.\left(\frac{x^2}{x+3}-1\right)+\frac{3x^2-14x+3}{x^2+3x}\)

\(A=\frac{\left(x^2-x-3\right)\left(-x^2+1\right)}{x\left(x+3\right)}+\frac{3x^2-14x+3}{x^2+3x}\)

\(A=\frac{\left(x^2-x-3\right)\left(1-x^3\right)}{\left(x+3\right)x}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)

\(A=\frac{\left(x^2-x-3\right)\left(1-x^2\right)+3x^2-14x+3}{\left(x+3\right)x}\)

\(A=\frac{-x^4+x^3+7x^2-15x}{x\left(x+3\right)}\)

\(A=\frac{x\left(-x^3+x^2+7x-15\right)}{x\left(x+3\right)}\)

\(A=\frac{-x^3+x^2+7x-15}{x+3}\)

\(A=-\frac{\left(x+3\right)\left(x^2-4x+5\right)}{x+3}\)

\(A=-\left(x^2-4x+5\right)\)

\(A=-x^2+4x-5\)

Trình độ hơi thấp, có gì sai sót xin bỏ qua cho :)

umk cảm ơn bạn trước nhé

1+1=2

:(((

1 + 1 = 2 

1 + 1 = 2 

ok chưa !! kick

ko ai rảnh viết hết đâu bn

b, (x-2)(x+1)^2 + (x+1)(x-2)^2 = 0 

(x-2)(x+1)[(x+1)+(x-2)]=0 

(x-2)(x+1)(2x-1)=0 

Therefore, three possible answers for x: 

(2x-1) = 0, x = 1/2 

(x+1) = 0, x = -1 

(x-2) = 0, x = 2 

X = 2, -1 or 1/2 

mk muốn mua bcs

\(\frac{7.x}{8}-5.\left(x-9\right)=20.x+\frac{1,5}{6}\)

\(\Leftrightarrow7.x-5.\left(x-9\right)=20x.8+\frac{1,5}{6}.8\)

\(\Leftrightarrow7.x-40\left(x-9\right)=160x+2\)

\(\Leftrightarrow7x-40+360=160x+2\)

\(\Leftrightarrow-33x+360=160x+2\)

\(\Leftrightarrow-33x=160x+x-360\)

\(\Leftrightarrow-33x=160x-358\)

\(\Leftrightarrow-33x-160x=160x-358\)

\(\Leftrightarrow-193x=-358\)

\(\Rightarrow x=\frac{358}{193}\)

Khong chac nhe

\(ĐKXĐ:\hept{\begin{cases}2x+y+1\ge0\\x+y\ge0\end{cases}}\)

Hệ \(\hept{\begin{cases}\sqrt{2x+y+1}-\sqrt{x+y}=1\\3x+2y=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x+y+1}=1+\sqrt{x+y}\\y=\frac{4-3x}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+y+1=1+2\sqrt{x+y}+x+y\\y=\frac{4-3x}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\sqrt{x+y}\\y=\frac{4-3x}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2=4x+4y\\y=\frac{4-3x}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\left(1\right)\\x^2=4x+4.\frac{4-3x}{2}\left(2\right)\\y=\frac{4-3x}{2}\left(3\right)\end{cases}}\)

Giải (2) :

\(\left(2\right)\Leftrightarrow x^2=4x+8-6x\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)

Kết hợp với (1) được x = 2

Thay x = 2 vào (3) được y = -1

Thấy x = 2 ; y = -1 t/m ĐKXĐ
Vậy hệ có nghiệm \(\hept{\begin{cases}x=2\\y=-1\end{cases}}\)