Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)88-3(7+x)=64\) \(c)[(x+32)-17]\div2=42\)
\(\Rightarrow3(7+x)=88-64\) \(\Rightarrow(x+32)-17=42.2\)\(\Rightarrow x+32-17=84\)
\(\Rightarrow3(7+x)=24\) \(\Rightarrow x=84+17-32\)
\(\Rightarrow7+x=24\div3\) \(\Rightarrow x=69\)
\(\Rightarrow7+x=8\)
\(\Rightarrow x=8-7\)
\(\Rightarrow x=1\)
\(b)131.x-941=2^7.2^3\) \(d)[(x^2+54)-32]\div2=61\)
\(\Rightarrow131.x-941=2^{10}\) \(\Rightarrow(x^2+54)-32=61.2\)
\(\Rightarrow131.x=1024+941\) \(\Rightarrow x+54-32=122\)
\(\Rightarrow131.x=1965\) \(\Rightarrow x=122+32-54\)
\(\Rightarrow x=1965\div131\) \(\Rightarrow x=100\)
\(\Rightarrow x=15\)
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: \(\Leftrightarrow x^3=\dfrac{539}{64}\)
hay \(x=\dfrac{7\sqrt{11}}{4}\)
c: \(\Leftrightarrow2^{2x-1}=2^9\cdot2^2=2^{11}\)
=>2x-1=11
hay x=6
d: \(\Leftrightarrow x^{17}-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
hay \(x\in\left\{0;1;-1\right\}\)
1a/ \(\left(15-x\right)+\left(x-12\right)=7-\left(-5+x\right)\)
=> \(\left(15-x\right)+\left(x-12\right)+\left(-5+x\right)=7\)
=> \(15-x+x-12-5+x=7\)
=> \(\left(15-12-5\right)-\left(x+x+x\right)=7\)
=> \(\left(15-12-5\right)-7=3x\)
=> \(3x=-2-7\)
=> \(3x=-9\)
=> \(x=\frac{-9}{3}=-3\)
b/ \(x-\left\{57-\left[42+\left(-23-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)
=> \(x-57-42-23-x=13-47+25-32+x\)
=> \(x-x+x=13-47+25-32+57+42+23\)
=> \(x=\left(13+23\right)-\left(47+57\right)+\left(25+57\right)-\left(32+42\right)\)
=> \(x=36-104+82-74\)
=> \(x=-60\)
d/ \(\left(x-3\right)\left(2y+1\right)=7\)
Vì 7 là số nguyên tố nên ta có 2 trường hợp:
TH1: \(\hept{\begin{cases}x-3=1\\2y+1=7\end{cases}}\)=> \(\hept{\begin{cases}x=4\\y=3\end{cases}}\).
TH2: \(\hept{\begin{cases}x-3=7\\2y+1=1\end{cases}}\)=> \(\hept{\begin{cases}x=10\\y=0\end{cases}}\).
Các cặp (x, y) thoả mãn điều kiện: \(\left(4;3\right),\left(10;0\right)\).
2x:2=32
==> 2x—1=25
==> x—1=5
x=5+1
x=6
5x—1:5=53
==> 5x—2=53
==> x—2=3
x—2=3
x=3+2
x=5
(2x—1)3=125
(2x—1)3=53
==> 2x—1=5
2x=5+1
2x=6
x=6:2
x=3
x17=x3
==>x=0 hoặc x=1
Mình quên cách lập luận bài này rồi bạn lên mạng tham khảo thêm nha
a) 2^x;2=32
Suy ra:2^x=32:2
Suy ra :2^x=16
Mà 16=2^4
Suy ra :x=4
Vậy x=4
Lát nữa mình giải nốt,bây giờ mình có việc.k cho mình nhé
a, 3x - 5 = 16...............
=> 3x = 16 + 5 = 21..................
=> x = 21 : 3 = 7..........
\(b,42-2\left(32-2^{x+1}\right)=10\)
\(\Rightarrow2\left(32-2^{x+1}\right)=42-10=32\)
\(\Rightarrow32-2^{x+1}=32:2=16\)
\(\Rightarrow2^{x+1}=32-16=16=2^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=4-1=3\)
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) \(\left[\left(x^2+54\right)-32\right]:2=61\)
\(\left(x^2+54\right)-32=61.2\)
\(\left(x^2+54\right)-32=122\)
\(x^2+54=122+32\)
\(x^2+54=154\)
\(x^2=154-54\)
\(x^2=100\)
\(x^2=10^2\)
\(x=10\)