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4 × 125 × 25 × 8
= (4 × 25) × (125 × 8)
= 100 × 1000
= 100 000
2 × 8 × 50 × 25 × 125
= (2 × 50) × (8 × 125) × 25
= 100 × 1000 × 125
= 12 500 000
2 × 3 × 4 × 5 × 50 × 25
= (2 × 50) × (4 × 25) × (3 × 5)
= 100 × 100 × 15
= 150 000
25 × 20 × 125 × 8 - 8 × 20 × 5 × 125
= 8 × 125 × 20 × (25 - 5)
= 1000 × 20 × 20
= 400 000
- (25 x 4) x (125 x8)=100 x 1000 =100000
- (50 x 2) x (8 x 125) x 25=100 x 1000 x 25 =2500000
- (2 x 50 ) x (4 x 25) x (3 x5)=100 x 100 x 15=150000
- 25 x 20 x125 x 8 - 5
= (20 x 5) x (125 x 8) -25=100 x 1000 -5 =100000 - 5 =99995
b){x-[52+(35-80.3)3-51]50+14}=20
=>x-[25+(243-240)3-51]50+14=20
=>x-[25+33-51]50+14=20
=>x-[25+27-51]50+14=20
=>x-[52-51]50+14=20
=>x-150+14=20
=>x-1+14=20
=>x-1=20-14
=>x-1=6
=>x=6+1
=>x=7
a){x-[52+(35-80.3)3-51]50+14}=20
=>x-[25+(243-240)3-51]50+14=20
=>x-[25+33-51]50+14=20
=>x-[25+27-51]50+14=20
=>x-[52-51]50+14=20
=>x-150+14=20
=>x-1+14=20
=>x-1=20-14
=>x-1=6
=>x=6+1
=>x=7
a) 84
b) -300
c) 1700
a) = 10 . ( 7.8 )
= 10 . 56
= 560
b)
= 9.2.7.5
= ( 9.7 ) . ( 2.5 )
= 63 . 10
= 630
b) 7.x=42
x = 42 : 7
x = 6
c) 180 : x = -12
x = 180 : ( -12 )
x = -15
Bài toán sai
bài toán đúng là x+(x+1)+(x+2)+(x+3)+...+19+20=20
`x^2=3`
`=>x=\sqrt{3}\or\x=-\sqrt{3}`
`x^2=36`
`<=>x^2=(+-6)^2`
`<=>x=+-6`
`x^2=25`
`<=>x^2=(+-5)^2`
`<=>x=+-5`
`2x^2+(-20)=55`
`<=>2x^2-20=55`
`<=>2x^2=75`
`<=>x^2=75/2`
`<=>x=+-\sqrt{75/2}`
`2(x-1)^2+5^0=9`
`<=>2(x-1)^2+1=9`
`<=>2(x-1)^2=8`
`<=>(x-1)^2=4`
`<=>x-1=2\or\x-1=-2`
`<=>x=3\or\x=-1`
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.
e: \(\Leftrightarrow x\inƯ\left(24\right)\)
mà x là số chẵn
nên \(x\in\left\{2;4;6;8;12;24;-2;-4;-6;-8;-12;-24\right\}\)
f: \(\Leftrightarrow x+1\inƯ\left(20\right)\)
\(\Leftrightarrow x+1\in\left\{1;2;4;5;10;20\right\}\)
\(\Leftrightarrow x\in\left\{0;1;3;4;9;19\right\}\)
mà 5<x<20
nên \(x\in\left\{9;19\right\}\)
h: \(x\inƯ\left(50\right)\)
nên \(x\in\left\{1;2;5;10;25;50\right\}\)
mà \(x\in B\left(25\right)\)
nên \(x\in\left\{25;50\right\}\)
50-5x(x-2)=20
5x(x-2)=50-20
5x(x-2)=30
(x-2)=30/5
(x-2)=6
x=6+2
x=8
50 - 5 x ( x -2 ) = 20
5 x ( x - 2) =50 -20
5 x ( x - 2 ) = 30
x - 2 = 30 : 5
x - 2 = 6
x = 6 + 2
x = 8
chúc bạn học tốt!