à em thêm nhầm số 1 vào

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Làm các bài tập trên màn chiếu vào vở nhé

a) \(\frac{2a^2-3a-2}{a^2-4}=2\)

\(\Rightarrow2a^2-3a-2=2\left(a^2-4\right)\)

\(\Rightarrow2a^2-3a-2=2a^2-4\)

\(\Rightarrow-3a-2=-4\)

\(\Rightarrow-3a=-2\Rightarrow a=\frac{2}{3}\)

b) \(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)

\(\Rightarrow\frac{\left(3a-1\right)\left(a+3\right)+\left(3a+1\right)\left(a-3\right)}{\left(3a+1\right)\left(a+3\right)}=2\)

\(\Rightarrow\frac{6a^2-6}{3a^2+10a+3}=2\)

\(\Rightarrow6a^2-6=2\left(3a^2+10a+3\right)\)

\(\Rightarrow6a^2-6=6a^2+20a+6\)

\(\Rightarrow-6=20a+6\Rightarrow20a=-12\)

\(\Rightarrow a=\frac{-3}{5}\)

\(\frac{x+10}{2000}+\frac{x+20}{1990}+\frac{x+30}{1980}+\frac{x+40}{1970}=-4\)

\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)

\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)

Vì  \(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}>0\)

\(\Rightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)

\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)

mà\(\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)\ne0\Rightarrow\left(x+2010\right)=0\\ \Rightarrow x=-2010\)

Ta có : (2x - 1)² - 25 = 0

=> (2x - 1)² = 25

=> \(\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

1)

\(27+(x-3)(x^2+3x+9)=-x\)

\(\Leftrightarrow 27+(x^3-3^3)=-x\)

\(\Leftrightarrow x^3=-x\)

\(\Leftrightarrow x^3+x=0\Leftrightarrow x(x^2+1)=0\)

\(\Rightarrow \left[\begin{matrix} x=0\\ x^2+1=0(vl)\end{matrix}\right.\)

Vậy $x=0$

2)

\(-4(x+2)-7(2x-1)+9(4-3x)=30\)

\(\Leftrightarrow -4x-8-14x+7+36-27x=30\)

\(\Leftrightarrow -45x+35=30\Leftrightarrow -45x=-5\)

\(\Rightarrow x=\frac{-5}{-45}=\frac{1}{9}\)

3)

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

\(\Leftrightarrow x^2-2.2x+2^2=0\)

\(\Leftrightarrow (x-2)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

4)

\((x-1)(x^2+x+1)-x(x+2)(x-2)=5\)

\(\Leftrightarrow (x^3-1^3)-x[(x+2)(x-2)]=5\)

\(\Leftrightarrow x^3-1-x(x^2-2^2)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow 4x-1=5\Rightarrow 4x=6\Rightarrow x=\frac{6}{4}=\frac{3}{2}\)