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( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0
<=> +) 1/7 . x - 2/7 = 0 +) (- 1 / 5) . x +3/5 = 0 +) 1/ 3 . x + 4/ 3 = 0
x = 2 x = 3 x = 4
Vậy x = 2 : x = 3 ; x=4
\(\frac{1}{20}\left(x-\frac{8}{15}\right)=-\frac{1}{30}\) \(\left(28+\frac{1}{5}\right).\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{1}{30}:\frac{1}{20}\) \(\frac{141}{5}.\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{2}{3}\) \(\frac{3}{5}.x+\frac{4}{7}=0\)
\(x=-\frac{2}{3}+\frac{8}{15}\) \(\frac{3}{5}.x=-\frac{4}{7}\)
\(x=-\frac{2}{15}\) \(x=-\frac{20}{21}\)
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)
e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)
g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)
h)Tương tự các câu trên
i) x = 0
k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)
l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)
=> x + 1 = 3 => x = 2
x.(x+1)=0
suy ra x=0 hoac x+1=0
x=0-1
x=-1
vay x=0 hoac x=-1
mấy câu sau cũng làm tương tự
1: \(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(4x+3\right)^2+8\left(x+3\right)^2\)
\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)
\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)
=56
2: \(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)
\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)
\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)
\(=6\)
