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a) \(58+7x=100\)
\(=>7x=100-58\)
\(=>7x=42\)
\(=>x=42:7\)
\(=>x=6\)
b) \(3x-7=28\)
\(=>3x=28+7\)
\(=>3x=35\)
\(=>x=35:3\)
\(=>x=\dfrac{35}{3}\)
c) \(x-56:4=16\)
\(=>x-14=16\)
\(=>x=16+14\)
\(=>x=30\)
d) \(101+\left(36-4x\right)=105\)
\(=>36-4x=105-101\)
\(=>36-4x=4\)
\(=>4x=36-4\)
\(=>4x=32\)
\(=>x=32:4\)
\(=>x=8\)
e) \(\left(x-12\right):12=12\)
\(=>x-12=12.12\)
\(=>x-12=144\)
\(=>x=144-12\)
\(=>x=132\)
f) \(\left(3x-2^4\right).7^3=2.7^4\)
\(=>3x-2^4=2.7^4:7^3\)
\(=>3x-16=2.7=14\)
\(=>3x=14+16\)
\(=>3x=30\)
\(=>x=30:3\)
\(=>x=10\)
i) \(\left(10+2x\right).4^{2011}=4^{2013}\)
\(=>10+2x=4^{2013}:4^{2011}\)
\(=>10+2x=4^2=16\)
\(=>2x=16-10\)
\(=>2x=6\)
\(=>x=6:2\)
\(=>x=3\)
\(#WendyDang\)
thanks
Bài 1 Tìm x biết:
a)65-(29-x)=32
65 -29+x=31
x=31-65+29
x=-5
b)(x+5)-(x+23)=x-34
x+5 -x +23 = x-34
(x-x)+ (23+5)=x-34
0+28=x-34
28=x-34
28+34=x
62=x
=>x=62
c)(16-x)+(x-38)=x+44
16-x+x-38=x+44
-x+x-x=44-16+38
-x=36
=>x=-36
d)-12+3(-x+7)=-18
3(-x+7)=-18+12
3(-x+7)=-6
-x+7=-6:3
-x+7=-2
-x=-2-7
-x=-9
=>x=9
Baif 2
d)|7-x|=10
=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)
e)(x-6).(7-2x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)
f)(9-x).(2x+8)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
g)x(-x+8).(-3x-18)=0
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)
h)(-x+8).(x-54).(-24-x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
a) Ta có: x + 34 là bội của x + 1
=> x + 34 \(⋮\)x + 1
=> (x + 1) + 33 \(⋮\)x + 1
=> 33 \(⋮\)x + 1
=> x + 1 \(\in\)Ư(33) = {1; 3; 11; 33}
Với: x + 1 = 1 => x = 1 - 1 = 0
x + 1 = 3 => x = 3 - 1 = 2
x + 1 = 11 => x = 11 - 1 = 10
x + 1 = 33 => x = 33 - 1 = 32
b) HD: 2x + 1 là ước của 4x + 82
=> 4x + 82 là B(2x + 1)
=> 2(2x + 1) + 80 \(⋮\)2x + 1
=> 80 \(⋮\)2x + 1 => 2x + 1 \(\in\)Ư(80) = {...} (tự liệt kê)
còn lại TT câu a
c) cx như trên
d) Ta có: 3x + 6 \(⋮\)2x - 1
=> 2(3x + 6) \(⋮\)2x - 1
=> 6x + 12 \(⋮\)2x - 1
=> 3(2x - 1) + 15 \(⋮\)2x - 1
=> 15 \(⋮\)2x - 1 => 2x - 1 \(\in\)Ư(15) = {1; 3; 5; 15}
còn lại tự lm (như trên)
g)\(2907\left(2x+1\right)=8721\)
⇔\(2x+1=3\)
⇔\(2x=2\)
⇔\(x=1\)
h)\(\left(4x-16\right):1905=60\)
⇔\(4x-16=114300\)
⇔\(4x=114316\)
⇔\(x=28579\)
i)\(23+3x=5^6:5^3\)
⇔\(23+3x=5^3\)
⇔\(23+3x=125\)
⇔\(3x=102\)
⇔\(x=34\)
k)\(219-7\left(x+1\right)=100\)
⇔\(7\left(x+1\right)=119\)
⇔\(x+1=17\)
⇔\(x=16\)