Tìm X:
a) X x 26 = 345
b) X : 16 = 946 + 345
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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) \(\left(12-12\dfrac{1}{3}\right):x+\dfrac{1}{6}=-\dfrac{2}{3}\)
\(-\dfrac{1}{3}x=-\dfrac{2}{3}-\dfrac{1}{6}\)
\(-\dfrac{1}{3}x=-\dfrac{5}{6}\)
\(x=-\dfrac{5}{6}:\left(-\dfrac{1}{3}\right)\)
\(x=\dfrac{5}{2}\)
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: Ta có: \(71-\left(x+33\right)=26\)
\(\Leftrightarrow x+33=45\)
hay x=12
b: Ta có: \(\left(x+73\right)-26=76\)
\(\Leftrightarrow x+73=102\)
hay x=29
c: Ta có: \(45-\left(x+9\right)=6\)
\(\Leftrightarrow x+9=39\)
hay x=30
a) 71 - ( 33 + x ) = 26
(33 + x ) = 71 -26
33 + x = 45
x = 45 - 33 = 12
b) ( x + 73 ) - 26 = 76
( x + 73 ) = 102
x = 102 - 73 = 29
c) 45 - ( x + 9 ) = 6
( x + 9 ) = 45 - 6 = 39
x = 39 - 9 = 30
d) 89 - ( 73 - x ) = 20
73 - x = 89 - 20
73 - x = 69
x = 73 - 69 = 4
e) 4(x+41) = 400
x + 41 = 400 : 4 = 100
x = 100 - 41 = 59
f) 11(x-9 ) = 77
x - 9 = 77 : 11 = 7
x = 7 + 9 = 16
g) x + 7 = 25 + 13 = 38
x = 38 - 7 = 31
h) x + 4 = 198 - 120 = 78
x = 78 - 4 = 74
i) x - 9 = 350 : 5 = 70
x = 70 + 9 = 79
j) 2x - 49 = 5 . 9 = 45
2x = 45 + 49 = 94
x = 94 : 2 = 47
k) 25 + 3( x - 8 ) = 106
3(x-8 ) = 106 - 25 = 81
x - 8 = 81 : 3 = 27
x = 27 +8= 35
l) 9( x + 4 ) - 25 = 20
9( x + 4 ) = 20 + 25 = 45
x + 4 =45 : 9 = 5
x = 5 - 4 = 1
m) 200 - ( 2x + 6 ) = 64
2x + 6 = 200 - 64 = 136
2x = 136 - 6 = 130
x = 130 : 2 = 65
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) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
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) X x 26 = 345 b) X : 16 = 946 + 345
X = 345 : 26 X : 16 = 1291
X = 13,26923077 X = 1291 x 16
X = 20656
/HT\
a)13,2692..... b)20656