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(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
\(1,x.\left(x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)
\(2,\left(x+12\right).\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)
\(3,\left(-x+5\right).\left(3-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Rightarrow\orbr{\begin{cases}-x=-5\\x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)
\(4,24:\left(3x-2\right)=-3\)
\(3x-2=-8\)
\(3x=-6\)
\(x=-2\)
\(5,-45:5\left(-3-2x\right)=3\)
\(5\left(-3-2x\right)=-15\)
\(-3-2x=-3\)
\(2x=0\)
\(x=0\)
\(6,x.\left(2+x\right)\left(7-x\right)=0\)
\(x=0\) hoặc \(\orbr{\begin{cases}2+x=0\\7-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=7\end{cases}}}\)
\(7,\left(x-1\right)\left(x+2\right)\left(-x+3\right)=0\)
TH1: x-1=0 TH2 : x+2=0 TH3: -x+3=0
x=1 x=-2 -x=-3 => x=3
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
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
1,
( 2x-5) + 17=6
\(2x-5=6-17\)
\(2x-5=-11\)
\(2x=-11+5\)
\(2x=-6\)
\(x=-6:2\)
\(x=-3\)
Vậy \(x=3\)
2,
10-2(4-3x)=-4
\(-2\left(4-3x\right)=-4-10\)
\(-2\left(4-3x\right)=-14\)
\(4-3x=-14:\left(-2\right)\)
\(4-3x=7\)
\(-3x=7-4\)
\(-3x=3\)
\(x=3:\left(-3\right)\)
\(x=-1\)
Vậy \(x=-1\)
3,
-12+3(-x+7)=-18
\(3\left(-x+7\right)=-18+12\)
\(3\left(-x+7\right)=-6\)
\(-x+7=-6:3\)
\(-x+7=-2\)
\(-x=-2-7\)
\(-x=-9\)
\(x=9\)
\(\text{Vậy }x=9\)
4,
24:(3x -2)=-3
\(3x-2=24:\left(-3\right)\)
\(3x-2=-8\)
\(3x=-8+2\)
\(3x=-6\)
\(x=-6:3\)
\(x=-2\)
\(\text{Vậy }x=-2\)
5,
-45:5.(-3-2x)=3
\(5\left(-3-2x\right)=-45:3\)
\(5\left(-3-2x\right)=-15\)
\(-3-2x=-15:5\)
\(-3-2x=-3\)
\(-2x=-3+3\)
\(-2x=0\)
\(x=0\)
Vậy \(x=0\)
6,
x.(x+7)= 0
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
\(\text{Vậy}\left\{{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
7,
( x+ 12 ) .( x-3)=0
\(\Rightarrow\left\{{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
8,
(-x+5).(3-x) =0
\(\Rightarrow\left\{{}\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-x=-5\\-x=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy \(x=5\text{ hoặc }x=3\)
9, x.( 2+x).(7-x)=0
\(\Rightarrow\left\{{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
10,
(x-1).(x+2).(-x-3)=0
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+2=0\\-x-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=-2\\-x=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
1) |x + 2| = 4
\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
2) 3 – |2x + 1| = (-5)
\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)
3) 12 + |3 – x| = 9
\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)
=>\(x=\varnothing\)
1) I x+2 I=4
\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)
2) \(3-|2x+1|=-5\)
\(\Leftrightarrow|2x+1|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)
3) \(12+|3-x|=9\)
\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)