Tìm x:
a)\(x\times8=16\); b)\(x\div3=9\);
c)\(4\times x=32\); d)\(50\div x=10\).
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Bài 2:
a: Ta có: \(2x+79=x+45\)
nên 2x-x=45-79
hay x=-24
b: Ta có: \(15-\left(x+7\right)=\left(x-8\right)+22\)
\(\Leftrightarrow8-x-x-14=0\)
\(\Leftrightarrow2x=22\)
hay x=11
\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
a) \(\Leftrightarrow x\left(x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) \(\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)
a) \(\left(12-12\dfrac{1}{3}\right):x+\dfrac{1}{6}=-\dfrac{2}{3}\)
\(-\dfrac{1}{3}x=-\dfrac{2}{3}-\dfrac{1}{6}\)
\(-\dfrac{1}{3}x=-\dfrac{5}{6}\)
\(x=-\dfrac{5}{6}:\left(-\dfrac{1}{3}\right)\)
\(x=\dfrac{5}{2}\)
b) \(\dfrac{4}{x}=\dfrac{x}{16}\)
\(x^2=4.16\)
\(x^2=64\)
\(\Rightarrow x=8;x=-8\)
`a)=>(12-37/3):x+1/6=-2/3`
`=>(12-37/3):x=-5/6`
`=>(-1/3):x=-5/6`
`=>x=(-1/3):(-5/6)`
`=>x=6/15=2/5`
`b)4/x=x/16`
`=>x^2=4*16`
`=>x^2=64`
`=>x^2=(+-8)^2`
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
a) ĐK: x ≥ 2
\(\sqrt{3x-6}=3\)
\(\Leftrightarrow3x-6=9\)
<=> 3x = 15
<=> x = 5
Vậy:....
b) ĐK: 5x - 16 ≥ 0
<=> 5x ≥ 16
<=> x ≥ 16/5
\(\sqrt{5x-16}=2\)
<=> 5x - 16 = 4
<=> 5x = 20
<=> x = 4
c) ĐK: \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
bình phương hai vế ta được:
a)điều kiện của x:x≥2
3x-6=9 <=> x=5(nhận)
b)ĐK: x≥16/5
5x-16=4 <=>x=4(nhận)
c) ta có: \(\dfrac{2x-3}{\left(x-2\right)^2-1}\)= \(\dfrac{2x-3}{\left(x-3\right)\left(x-1\right)}\)
ĐKXĐ: x≠3 ;x≠1
`#3107`
a)
`8x^2 - 5 = 67 `
`=> 8x^2 = 67 + 5`
`=> 8x^2 = 72`
`=> x^2 = 9`
`=> x^2 = (+-3)^2`
`=> x = +-3`
Vậy, `x \in {-3;3}`
b)
`(4x - 2)^4=16`
`=> (4x - 2) = (+-2)^4`
`=> ` TH1: `4x - 2 = 2`
`=> 4x = 4`
`=> x =1`
TH2: `4x - 2 = -2`
`=> 4x = 0`
`=> x=0`
Vậy, `x \in {0; 1}.`
a) \(8x^2-5=67\)
\(8x^2=72\)
\(x^2=9\)
\(x=3\)
Vậy x = 3
b) \(\left(4x-2\right)^4=16\)
\(\left(4x-2\right)^2=4\)
\(4x-2=2\)
\(4x=4\)
\(x=1\)
Vậy x = 1
a) x = 2 .
b) x = 27 .
c) x = 8 .
d) x = 5 .
a) x=2
b) x =27
c) x=8
d) x= 5