bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a: \(A=4x-3x^2+20-15x-9x^2-12x-4+\left(2x+1\right)^3-\left(8x^3-1\right)\)

\(=-12x^2-23x+16+8x^3+12x^2+6x+1-8x^3+1\)

\(=-17x+18\)

\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}=\frac{x+2}{x+2}+\frac{-5}{x^2+x-6}+\frac{-1}{x-2}\)

=\(\frac{\left(x+2\right)\left(x-2\right)}{x^2+x-6}+\frac{-5}{x^2+x-6}+\frac{-1\left(x+3\right)}{x^2+x-6}=\frac{\left(x+2\right)\left(x-2\right)-5-1\left(x+3\right)}{x^2+x-6}\)

=\(\frac{x^2-4-5-x-3}{x^2+x-6}=\frac{x^2-x-12}{x^2+x+6}\)

\(\frac{x^2-x-12}{x^2+x-6}=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)

Để giá trị của PT A được xác định thì \(\left(x-2\right)\ne0\)và \(\left(x+3\right)\ne0\)

=> \(x\ne2\) và \(x\ne-3\) thì PT được xác định

@__@ Lag cả cái đề

b) \(\frac{x^2-x-12}{x^2+x-6}=\frac{x^2+3x-4x-12}{\left(x-2\right)\left(x+3\right)}=\frac{x\left(x+3\right)-4\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)

giúp mình mọi người ơi

a: Thay x=49 vào A, ta được:

\(A=\dfrac{2\cdot7+1}{7-3}=\dfrac{14+1}{4}=\dfrac{15}{4}\)

b: \(B=\dfrac{2x+36}{x-9}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36-9\left(\sqrt{x}+3\right)-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x+36-9\sqrt{x}-27-x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\)

P>1 khi P-1>0

=>\(\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}>0\)

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

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

=>x>4

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>4\\x\ne9\end{matrix}\right.\)

yggucbsgfuyvfbsudy

????????

a, \(A=x^2\left(2x-1\right)+x\left(x+8\right)=2x^3-x^2+x^2+8x=2x^3+8x\)

Thay x = -2, ta có:

\(2\cdot\left(-2\right)^3+8\cdot\left(-2\right)=-32\)

b, \(A=2x^3+8x=0\\ \Leftrightarrow2x\left(x^2+4\right)=0\\ \Leftrightarrow x=0\)

Vậy A=0 khi x=0

a,A = \(x^2\).( 2\(x\) - 1) + \(x\)(\(x+8\))

A = 2\(x^3\) - \(x^2\) + \(x^2\) + 8\(x\)

A = 2\(x^3\) + 8\(x\)

b, \(x=-2\) ⇒ A = 2.(-2)³ + 8.(-2) = - 32 

A = 0 ⇔ 2\(x^3\) + 8\(x\) = 0

             2\(x\left(x^2+4\right)\) = 0 

              vì \(x^2\) + 4 > 0 ∀ \(x\) ⇒ \(x\)  =0 

\(a,A=x^2\left(2x-1\right)+x\left(x+8\right)\\ =2x^3-x^2+x^2+8x=2x^3+8x\\ =2x.\left(x^2+8\right)\\ Thế.x=-2.vào.A:A=2x.\left(x^2+8\right)=2.\left(-2\right).\left[\left(-2\right)^2+8\right]=-4.12=-48\\ b,A=0\\ \Leftrightarrow2x\left(x^2+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+8=0\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=0\\ Vậy:A=0\Leftrightarrow x=0\)

a)\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

A xác định

\(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\\left(x+3\right)\left(x-2\right)\ne0\\x\ne2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

Vậy A xác định \(\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

b) \(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x^2-2x\right)+\left(3x-6\right)}+\frac{1}{2-x}\)

\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x.\left(x-2\right)+3.\left(x-2\right)}+\frac{1}{2-x}\)

\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{1}{2-x}\)

\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)

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

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

\(A=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

\(A=\frac{\left(x^2+3x\right)-\left(4x+12\right)}{\left(x+3\right)\left(x-2\right)}\)

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

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

\(A=\frac{x-4}{x-2}\left(x+3\ne0\right)\)

c) \(A=-\frac{3}{4}\)

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

\(\Leftrightarrow4.\left(x-4\right)=-3.\left(x-2\right)\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow7x=22\)

\(\Leftrightarrow x=\frac{22}{7}\)

Vậy \(x=\frac{22}{7}\)

Tham khảo nhé~