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\(a)88-3(7+x)=64\) \(c)[(x+32)-17]\div2=42\)
\(\Rightarrow3(7+x)=88-64\) \(\Rightarrow(x+32)-17=42.2\)\(\Rightarrow x+32-17=84\)
\(\Rightarrow3(7+x)=24\) \(\Rightarrow x=84+17-32\)
\(\Rightarrow7+x=24\div3\) \(\Rightarrow x=69\)
\(\Rightarrow7+x=8\)
\(\Rightarrow x=8-7\)
\(\Rightarrow x=1\)
\(b)131.x-941=2^7.2^3\) \(d)[(x^2+54)-32]\div2=61\)
\(\Rightarrow131.x-941=2^{10}\) \(\Rightarrow(x^2+54)-32=61.2\)
\(\Rightarrow131.x=1024+941\) \(\Rightarrow x+54-32=122\)
\(\Rightarrow131.x=1965\) \(\Rightarrow x=122+32-54\)
\(\Rightarrow x=1965\div131\) \(\Rightarrow x=100\)
\(\Rightarrow x=15\)
A) 88-3(7+x)=64
3(7+x)=88-64
3(7+x)=24
7+x=24:3
7+x=8
X=8-7
X=1
a) => (x+32) - 17 = 42:2 = 21
=> x+32 = 21+17 = 38
=> x=38-32=6
b) => 61+(53-x) = 1785:17=105
=> 53-x = 105-61=44
=> x = 53-44 =9
c) => x^2 +54 -32 = 244:2 = 122
=> x^2 +22 = 122
=> x^2 = 122-2=100
=> x= 10 hoặc -10
giải oy pn **** giùm mk nka
a) x – 32 : 16 = 48 ó x – 2 = 48 ó x = 48 + 2 ó x = 50
b) 88 – 3.(7+x) = 64 ó 3.(7+x) = 88 – 64 ó 7 + x = 24:3 ó x = 8 – 7 ó x = 1
c) (5+4x) : 3 – 121 : 11 = 4 ó (5+4x) : 3 – 11 = 4 ó (5+4x) : 3 = 4 + 11 ó 5+4x = 15.3 ó 4x = 45 – 5 ó 4x = 40 ó x = 10
d) 15 – 2(3x+1) = 11.13 – 130 ó 15 – 2(3x+1) = 143 – 130 ó 15 – 2(3x+1) = 13
ó 2(3x+1) = 15 – 13 ó 3x + 1 = 2:2 ó 3x = 1 – 1 ó 3x = 0 ó x = 0
( ( x . 32 ) - 17 ) . 2 = 42
( ( x . 32 ) - 17 ) = 42 : 2
( ( x . 32 ) - 17 ) = 21
x . 32 = 21 + 17
x . 32 = 38
x = 38 : 32
x = 19/16
a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
Bài 1
a) 76 x 54 + 2 x 27 x 23 + 54 = 76 x 54 + 54 x 23 + 54 = 54 x (76 + 23 + 1) = 54 x 100 = 5400
b) 17 x 82 - 17 x 32 = 17 x (82 - 32) = 17 x 50 = 850
Bài 2
28 - (x + 3) = 14
x + 3 = 28 - 14
x + 3 = 14
x = 14 - 3
x = 11
a) \(88-3\cdot\left(7+x\right)=64\)
\(3\cdot\left(7+x\right)=88-64=24\)
\(7+x=\frac{24}{3}=8\)
\(x=8-7=1\)
b) \(\left[\left(x+32\right)-17\right]\cdot2=42\)
\(\left(x+32\right)-17=\frac{42}{2}=21\)
\(x+32=21+17=38\)
\(x=38-32=6\)
c) \(\left[\left(x^2+54\right)-32\right]:2=61\)
\(\left(x^2+54\right)-32=61\cdot2=122\)
\(x^2+54=122+32=154\)
\(x^2=154-54=100\)
\(\Rightarrow x=\sqrt{100}=10\)
a) \(88-3.\left(7+x\right)=64\)
\(3.\left(7+x\right)=88-64\)
\(21+3x=24\)
\(3x=3\)
\(x=1\)
b) \(\left[\left(x+32\right)-17\right].2=42\)
\(\left(x+32\right)-17=42:2\)
\(\left(x+32\right)-17=21\)
\(x+32=21+17\)
\(x+32=38\)
\(x=38-32=6\)
c) \(\left[\left(x^2+54\right)-32\right]:2=61\)
\(\left(x^2+54\right)-32=61.2\)
\(\left(x^2+54\right)-32=122\)
\(x^2+54=122+32\)
\(x^2+54=154\)
\(x^2=154-54\)
\(x^2=100\)
\(x^2=10^2\)
\(x=10\)