tìm x:
a) 2x : 25 = 4.
b) 4(x + 41) = 400
c) (x + 7) – 25 = 13
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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
`(13+x)/20 = 3/4`
`(13+x) xx4=3xx20`
`(13+x)xx4=60`
`13+x=60:4`
`13+x=15`
`x=15-13`
`x=2`
__
`(23-x)/25 =4/5`
`(23-x)xx5=4xx25`
`(23-x)xx5=100`
`23-x=100:5`
`23-x=20`
`x=23-20`
`x=3`
a)
⇔ \(x^2-16=9\)
⇔ \(x^2=25\)
⇔ \(x=\pm5\)
b)
⇔ \(x^2-4x+4-25x^2+20x-4=0\)
⇔ \(16x-24x^2=0\)
⇔ \(8x\left(2-3x\right)=0\)
⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)
c)
⇔ \(3x^2-10x-20=0\)
⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)
⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)
⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)
Vậy...
d)
⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)
⇔ 7x = 49
⇔ x=7
Vậy...
a. 4.(x+41) = 7
x + 41 = 7 : 4 = 1,75
x = 1,75 - 41 = -39,25
b. 4.(x-3) = 72 - 110 = 49 - 1 = 48
x - 3 = 48 : 4 = 12
x = 12 + 3 = 15
a) \(4\left(x+41\right)=400\)
\(\Rightarrow x+41=400:4\)
\(\Rightarrow x+41=100\)
\(\Rightarrow x=100-41\)
\(\Rightarrow x=59\)
2*x:25=4
2*x=4*25
2.x=100
x=100/2
x=50
a)x=50
b)x+59
c)x=31