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\(3x\left(x-1\right)+5\left(2-x\right)=3x^2-7x+6\) \(6\)
<=> \(3x^2-3x+10-5x=3x^2-7x+6\)
<=> \(-x=-4\)
<=> \(x=4\)
\(\left(x+2\right)^2=\frac{1}{2}-\frac{1}{3}\)
<=> \(\left(x+2\right)^2=\frac{1}{6}\)
<=> \(\hept{\begin{cases}x+2=\sqrt{\frac{1}{6}}\\x+2=-\sqrt{\frac{1}{6}}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\sqrt{\frac{1}{6}}-2\\x=-\sqrt{\frac{1}{6}}-2\end{cases}}\)
a: P(x)=0
=>4x-1/2=0
=>x=1/8
b: Q(x)=0
=>(x-1)(x+1)=0
=>x=1 hoặc x=-1
c: A(x)=0
=>-12x+18=0
=>-12x=-18
hay x=3/2
d: B(x)=0
=>-x2=-16
=>x=4 hoặc x=-4
e: C(x)=0
=>3x2=-12
=>\(x\in\varnothing\)
a) \(-312+13×\left(\:x-1\right)=\:-113\div213\)
\(-312 +13×\left(x-1\right)=-\frac{113}{213}\)
\(13×\:\left(x-1\right)=-\frac{113}{213}+312\)
\(13×\left(x-1\right)=311\)
\(x-1=311\div13\)
\(x-1=\frac{311}{13}\)
\(x=\frac{311}{13}+1\)
\(x=\frac{324}{13}\)
Vậy, \(x =\frac{324}{13}\)
Cbht
b) \(x-14=x-25\)
\(x-x =14-25\)
\(0= -11\)
=> x không tồn tại
Cbht
a) \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\Rightarrow\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{10}{12}\Rightarrow\dfrac{x}{12}=\dfrac{11}{12}\Rightarrow x=11\)
b) \(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\Rightarrow\dfrac{10}{15}-\dfrac{19}{15}x=\dfrac{-3}{5}\Rightarrow\dfrac{-19}{15}x=\dfrac{-13}{15}\Rightarrow x=\dfrac{13}{19}\)
c) \(\dfrac{\left(-3\right)^x}{81}=-27\Rightarrow\left(-3\right)^x=-2187\Rightarrow x=7\)
d) \(2^{x-1}=16\Rightarrow x-1=4\Rightarrow x=5\)
e) \(\left(x-1\right)^2=25\Rightarrow x-1=5\Rightarrow x=6\)
g) \(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\Rightarrow x=\dfrac{1}{12}\\x+\dfrac{1}{2}=0\Rightarrow x=\dfrac{-1}{2}\end{matrix}\right.\)
1, \(x^2-4x-4x+16=0\)
\(\Leftrightarrow x^2-8x+16=0\)
\(\Leftrightarrow\left(x-4\right)^2=0\)
\(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy.............
2, \(x^2+3x-5x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy...............
3, \(x^2-6x+8=0\)
\(\Leftrightarrow x^2-6x+9-1=0\)
\(\Leftrightarrow\left(x-3\right)^2-1=0\)
\(\Leftrightarrow\left(x-3\right)^3=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy......................
4, \(x^2+8x+12=0\)
\(\Leftrightarrow x^2+8x+16-4=0\)
\(\Leftrightarrow\left(x+4\right)^2-4=0\)
\(\Leftrightarrow\left(x+4\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=2\\x+4=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-6\end{matrix}\right.\)
Vậy............
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
giúp mình nha
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