tính giá trị của bt
\(C=\frac{3x^3-x^2-3x+2021}{3x^4-x^3+3x+2021}\) với \(x=\frac{1}{3}\)
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12 tháng 1 2019
a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2x+1-3x-1+x^2}{3x}\)
\(A=\frac{x^2-x}{3x}\)
\(A=\frac{x\left(x-1\right)}{3x}\)
\(A=\frac{x-1}{3}\)
b) Thay x = 4 ta có :
\(A=\frac{4-1}{3}=\frac{3}{3}=1\)
c) Để A thuộc Z thì \(x-1⋮3\)
\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)
\(\Rightarrow x\in\left\{1;4;7;...\right\}\)
Vậy.....
7 tháng 9 2021
a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)
13 tháng 4 2020
a) Ta có: \(\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)=\left(4-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)-\left(x-4\right)\right]-2\left(3x-2\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3-x+4\right)-6x+4=0\)
\(\Leftrightarrow x-4-6x+4=0\)
\(\Leftrightarrow-5x=0\)
mà -5<0
nên x=0
Vậy: x=0
14 tháng 7 2021
\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
31 tháng 12 2021
\(a,\left(3x+1\right)\left(3x-1\right)-\left(18x^3+5x^2-2x\right):2x\\ =\left(9x^2-1\right)-\left(9x^2+\dfrac{5}{2}x-1\right)\\ =9x^2-1-9x^2-\dfrac{5}{2}x+1=\dfrac{5}{2}x\)
\(b,3x\left(x-2021\right)-x+2021=0\\ \Rightarrow b,3x\left(x-2021\right)-\left(x-2021\right)=0\\ \Rightarrow\left(x-2021\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)
21 tháng 7 2018
b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)
\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)
\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)
\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)
Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)
YN
1 tháng 10 2021
Tìm giá trị nhỏ nhất của biểu thức:
\(P=3x^2+31y^2-18xy+6x-14y+2021\)
\(=3[\left(x^2-6xy+9y^2\right)+2\left(x-3y\right)+1]+\left(4y^2+4y+1\right)+2017\)
\(=3[\left(x-3y\right)^2+2\left(x-3y\right)+1]+\left(2y+1\right)^2+2017\)
\(=3\left(x-3y+1\right)^2+\left(2y+1\right)^2+2017\ge2017\)
Vậy \(MinP=2017\) khi \(\hept{\begin{cases}x-3y+1=0\\2y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{-5}{2}\\y=\frac{-1}{2}\end{cases}}\)
Thực hiện phép tính:
x^2 - x + 1 3x^2 - 2x + 2 3x^4 - 5x^3 + 7x^2 - 4x + 2 - 3x^4 - 3x^3 + 3x^2 -2x^3 + 4x^2 - 4x + 2 - -2x^3 + 2x^2 - 2x 2x^2 - 2x + 2 2x^2 - 2x + 2 0
TA CÓ :\(\frac{3x^3-x^2-3x+2021}{3x^{4^{ }}-x^3+3x+2021}\)= \(\frac{\left(x^2-1\right)\left(3x-1\right)+2020}{\left(3x-1\right)\left(x^3+1\right)+2022}\)
THAY x=1/3 vào ta được :\(\frac{\left(3.\frac{1}{3}-1\right)\left(\frac{1}{3^2}-1\right)+2020}{\left(3.\frac{1}{3}-1\right)\left(\frac{1}{3^3}+1\right)+2022}\)= \(\frac{2020}{2022}\)=\(\frac{1010}{1011}\)
Vậy C=1010/1011