2/3 + 1/3 : 3x = 20%
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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.
a) \(\left(3x-2\right)\left(3x+2\right)-\left(3x+4\right)^2=20\\ \Rightarrow9x^2-4-9x^2-24x-16-20=0\\ \Rightarrow-24x-40=0\\ \Rightarrow-24x=40\\ \Rightarrow x=-\dfrac{5}{3}\)
b) \(6x^2-2x\left(3x+1\right)=10\\ \Rightarrow6x^2-6x^2-2x=10\\ \Rightarrow-2x=10\\ \Rightarrow x=-5\)
c) \(x^2+4x+3=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
a ) 10 x X - 1 - 3 - 5 - 7 - ... - 19 = 2 + 4 + 6 + ... + 20
10 x X - 1 - 3 - 5 - 7 - ... - 19 = 110
10 x X - ( 1 + 3 + 5 + 7 + ... + 19 ) = 110
10 x X - 100 = 110
10 x X = 110 + 100
10 x X = 210
X = 210 : 10
X = 21
a) x2 - x - 12
= x2 - 4x + 3x - 12
= x(x - 4) + 3(x - 4)
= (x - 4)(x + 3)
b) x3 - y3 - 3x2 + 3x - 1
= (x3 - 3x2 + 3x - 1) - y3
= (x - 1)3 - y3
= (x - 1 - y) [ (x - 1)2 + (x - 1)y + y2 ]
= (x - y - 1)(x2 - 2x + 1 + xy - y + y2 )
d) 4x3 - 5x2 - 16x + 20
= (4x3 - 8x2) + (3x2 - 6x) - (10x - 20)
= 4x2 (x - 2) + 3x(x - 2) - 10(x - 2)
= (x - 2)(4x2 + 3x - 10)
= (x - 2)(4x2 + 8x - 5x - 10)
= (x - 2)(x + 2)(4x - 5)
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
b: \(=\left(x^2+3x+1-3x+1\right)^2=\left(x^2+2\right)^2\)
Tk mình đi mọi người mình bị âm nè!
Ai tk mình mình tk lại cho!!
a) \(-\dfrac{11}{20}\)
b) \(x=-\dfrac{13}{3}\)
c) \(x=\dfrac{22}{7}\)
d) \(x=-\dfrac{52}{3}\)
c: \(\dfrac{3}{7}-\dfrac{2}{3}x=\dfrac{-5}{3}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{3}{7}+\dfrac{5}{3}=\dfrac{44}{21}\)
\(\Leftrightarrow x=\dfrac{44}{21}:\dfrac{2}{3}=\dfrac{44}{21}\cdot\dfrac{3}{2}=22\cdot\dfrac{1}{7}=\dfrac{22}{7}\)
\(2\left(x-1\right)+3\left(3x-2\right)=x-4\)
\(2x-2+9x-6=x-4\)
\(2x+9x-x-2-6=-4\)
\(10x-2-6=-4\)
\(10x-2=2\)
\(10x=4\)
\(x=\frac{2}{5}\)
Vậy \(x=\frac{2}{5}\)
\(3\left(4-x\right)-2\left(x-1\right)=x+20\)
\(12-3x-2x+2=x+20\)
\(12-5x+2=x+20\)
\(12-5x-x+2=20\)
\(12-6x+2=20\)
\(12-6x=18\)
\(6x=-6\)
\(x=-1\)
Vậy \(x=-1.\)
\(4\left(2x+7\right)-3\left(3x-2\right)=24\)
\(8x+28-9x+6=24\)
\(8x-9x+28+6=24\)
\(-x+34=24\)
\(-x=-10\)
\(x=10\)
Vậy \(x=10\)
\(3\left(x-2\right)+2x=10\)
\(3x-6+2x=10\)
\(3x+2x-6=10\)
\(5x=16\)
\(x=\frac{16}{5}\)
Vậy \(x=\frac{16}{5}\)
2(x-1)+3(3x-2)=x-4
=>2x-2=9x-6-x+4=0
=>10x-4=0
=>x=\(\frac{2}{5}\)
\(\dfrac{2}{3}+\dfrac{1}{3}:3x=20\%\)
\(\dfrac{2}{3}+\dfrac{1}{3}\times\dfrac{1}{3x}=\dfrac{20}{100}\)
\(\dfrac{2}{3}+\dfrac{1}{9x}=\dfrac{1}{5}\)
\(\dfrac{1}{9x}=\dfrac{1}{5}-\dfrac{2}{3}\)
\(\dfrac{1}{9x}=\dfrac{3}{15}-\dfrac{10}{15}\)
\(\dfrac{1}{9x}=-\dfrac{7}{15}\)
\(9x\times\left(-7\right)=15\)
\(9x=\dfrac{15}{\left(-7\right)}\)
\(x=-\dfrac{15}{7}\times\dfrac{1}{9}\)
\(x=-\dfrac{5}{21}\)
\(\dfrac{2}{3}+\dfrac{1}{3}:3x=20\%\)
\(\Rightarrow\dfrac{1}{3}\cdot\dfrac{1}{3x}=\dfrac{1}{5}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{9x}=-\dfrac{7}{15}\)
\(\Rightarrow9x=-\dfrac{15}{7}\)
\(\Rightarrow x=-\dfrac{15}{7}:9\)
\(\Rightarrow x=-\dfrac{5}{21}\)