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Ta có: \(\left(x-1\right)^{2020}\ge0\forall x\)
\(\left|y-3\right|\ge0\forall y\)
Do đó: \(\left(x-1\right)^{2020}+\left|y-3\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy: (x,y)=(1;3)
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(2x - 1)³ - 8x + 4 = 0
(2x - 1)³ - 4x(2x - 1) = 0
(2x - 1)[(2x - 1)² - 4x] = 0
(2x - 1)[(2x - 1)(2x - 1) - 4x] = 0
(2x - 1)[2x(2x - 1) - 1.(2x - 1) - 4x] = 0
(2x - 1)(4x² - 2x - 2x + 1 - 4x) = 0
(2x - 1)(4x² + 1) = 0
⇒ 2x - 1 = 0 hoặc 4x² + 1 = 0
*) 2x - 1 = 0
2x = 1
x = 1/2
*) 4x² + 1 = 0
4x² = -1 (vô lý vì 4x² ≥ 0 với mọi x)
Vậy x = 1/2
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\(a,1-\left(\dfrac{\dfrac{5}{3}}{8}+x-\dfrac{\dfrac{7}{5}}{24}\right)-\dfrac{\dfrac{16}{2}}{3}=0\\ \Leftrightarrow1-\left(\dfrac{5}{24}+x-\dfrac{7}{120}\right)=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{3}{20}+x=1-\dfrac{8}{3}=-\dfrac{5}{3}\\ \Leftrightarrow x=-\dfrac{5}{3}-\dfrac{3}{20}=-\dfrac{109}{60}\)
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a) \(\left(\frac{1}{7}x-\frac{2}{3}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x-\frac{2}{3}=0\\-\frac{1}{5}x+\frac{3}{5}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x=\frac{2}{3}\\-\frac{1}{5}x=-\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{14}{3}\\x=3\end{cases}}\)
b)\(\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(\Leftrightarrow x.\left(\frac{1}{10}-\frac{4}{5}\right)+1=0\)
\(\Rightarrow-\frac{7}{10}x=-1\)
\(\Rightarrow x=\frac{10}{7}\)
c)\(\left(2x-\frac{1}{3}\right).\left(5x+\frac{2}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\5x+\frac{2}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\5x=-\frac{2}{7}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{2}{35}\end{cases}}\)
a, (1/7 . x - 2/3) . (-1/5 . x + 3/5) = 0
Suy ra : 1/7 .x -2/3 = 0 hoặc -1/5 .x + 3/5 =0
Vậy : 1/7 .x = 2/3 hoặc -1/5 .x = 3/5
x =2/3 : 1/7 hoặc x = 3/5 : (-1/5)
x = 14/3 hoặc x = -3
b, 1/10 .x - 4/5 .x + 1 =0
x . (1/10 - 4/5) + 1 = 0
x . (-7/10) + 1 = 0
x . -7/10 =0 +1 = 1
x = 1 : (-7/10)
x = -10/7
c, (2x - 1/3 ) . (5x +2/7) = 0
Suy ra : 2x - 1/3 = 0 hoặc 5x + 2/7 = 0
Vậy : 2x = 1/3 hoặc 5x = 2/7
x = 1/3 : 2 hoặc x = 2/7 : 5
x = 1/6 hoặc x = 2/35
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Hằng đẳng thức đó bn:
\(\left(a+b\right)\left(a^2-ab+b^2\right)\)
Thay vào thì: \(-\left(x-3\right)\left(x^2-3x+9\right)=-\left[\left(x-3\right)\left(x^2-3x+3^2\right)\right]\)
\(=-\left(x^3-27\right)=-x^3+27\)
Bài làm:
Ta có: \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)=\left(x-3\right)^3+3\left(2x+1\right)^2-\left(x^3-5x+1\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3+27=x^3-9x^2+27x-27+12x^2+12x+3-x^3+5x-1\)
\(\Leftrightarrow6x^2+41x-51=0\)
\(\Leftrightarrow6\left(x^2+\frac{41}{6}x+\frac{1681}{144}\right)-\frac{2905}{24}=0\)
\(\Leftrightarrow\left(x+\frac{41}{12}\right)^2-\frac{\left(\sqrt{2905}\right)^2}{12^2}=0\)
\(\Leftrightarrow\left(x+\frac{41}{12}-\frac{\sqrt{2905}}{12}\right)\left(x+\frac{41}{12}+\frac{\sqrt{2905}}{12}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{2905}-41}{12}\\x=\frac{-\sqrt{2905}-41}{12}\end{cases}}\)
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a, \(A\left(x\right)+4x^3-x=-5x^2-2x^3+5+3x^2+2x\\ \Leftrightarrow A\left(x\right)=-5x^2-2x^3+5+3x^2+2x-4x^3+x=\left(-2x^3-4x^3\right)+\left(-5x^2+3x^2\right)+\left(2x+x\right)+5\\ =-6x^3-2x^2+3x+5\)
b, \(B\left(x\right)=A\left(x\right):\left(x-1\right)=\left(-6x^3-2x^2+3x+5\right):\left(x-1\right)=-6x^2-8x-5\)
Thay \(x=-1\) vào \(B\left(x\right)\)
\(\Rightarrow-6.\left(-1\right)^2-8\left(-1\right)-5=-3\ne0\)
\(\Rightarrow x=-1\) không là nghiệm của B(x)
Cho từng cái = 0 rồi quất thôi b :)
\(\Leftrightarrow\hept{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}5x=0+1\\2x=0+\frac{1}{3}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=1:5\\x=\frac{1}{3}:2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x\in\phi\\2x=\frac{1}{6}\end{cases}}\)