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Đề bài ko đúng em, tử số bên trái là 32 mới hợp lý chứ không phải 3.2
Ta có: \(\left|5x+7\right|+\left|5x-1\right|=\left|5x+7\right|+\left|1-5x\right|\ge\left|5x+7+1-5x\right|=8\) (1)
\(\left(2y+1\right)^{2020}\ge0\Rightarrow3\left(2y+1\right)^{2020}+4\ge4\)
\(\Rightarrow\dfrac{32}{3\left(2y+1\right)^{2020}+4}\le\dfrac{32}{4}=8\) (2)
Từ (1); (2) \(\Rightarrow\left|5x+7\right|+\left|5x-1\right|\ge\dfrac{32}{3\left(2y+1\right)^{2020}+4}\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left(5x+7\right)\left(1-5x\right)\ge0\\2y+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{5}\le x\le\dfrac{1}{5}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
a,thay x=1,y=-1
=>A=(15.1+2.-1)-[(2.1+3)-(5.1+-1)]=13-[5-4]=12
b,thay=-1/2,y=1/7
=>B=4
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
1, \(\left|\frac{3}{2}x-1\right|-2x=1\Rightarrow\left|\frac{3}{2}x-1\right|=1+2x\)
Vì \(\left|\frac{3}{2}x-1\right|\ge0\Leftrightarrow1+2x\ge0\Leftrightarrow x\ge\frac{-1}{2}\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{2}x-1=1+2x\\\frac{3}{2}x-1=-1-2x\end{cases}\Rightarrow\orbr{\begin{cases}\frac{3}{2}x-2x=1+1\\\frac{3}{2}x+2x=-1+1\end{cases}\Rightarrow}\orbr{\begin{cases}\frac{-1}{2}x=2\\\frac{7}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-4\left(ktm\right)\\x=0\left(tm\right)\end{cases}}}\)
Vậy x = 0
2,3 tương tự 1
4, Vì \(\left|x\left(x^2-\frac{5}{4}\right)\right|\ge0\Rightarrow x\ge0\)
Ta có: \(\left|x\left(x^2-\frac{5}{4}\right)\right|=x\Rightarrow x\left(x^2-\frac{5}{4}\right)=\pm x\) (1)
- Nếu x = 0 thì 0 = 0 thỏa mãn (1)
- Nếu \(x\ne0\) thì \(\left(1\right)\Leftrightarrow\orbr{\begin{cases}x^2-\frac{5}{4}=1\\x^2-\frac{5}{4}=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=\frac{9}{4}\\x^2=\frac{1}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\pm\frac{3}{2}\\x=\pm\frac{1}{2}\end{cases}}}\)
Vì \(x\ge0\Rightarrow x\in\left\{0;\frac{1}{2};\frac{3}{2}\right\}\)
Vậy...
\(.a.\)
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Leftrightarrow\left[\begin{matrix}\left(x-7\right)^{x+1}=0\\\left[1-\left(x-7\right)^{10}\right]=0\end{matrix}\right.\)
+ Nếu \(\left(x-7\right)^{x+1}=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=0+7\)
\(\Rightarrow x=7\)
+ Nếu \(1-\left(x-7\right)^{10}=0\)
\(\Rightarrow\left(x-7\right)^{10}=1\)
\(\Rightarrow\left(x-7\right)^{10}=\left(\pm1\right)^{10}\)
\(\Rightarrow\left[\begin{matrix}x-7=1\\x-7=-1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=1+7\\x=-1+7\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=8\\x=6\end{matrix}\right.\)
Vậy : \(x\in\left\{6;7;8\right\}\)
a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)