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HP
14 tháng 8 2021
\(\dfrac{2x-3}{5}-x+2\ge\dfrac{x}{3}\)
\(\Leftrightarrow3\left(2x-3\right)-15\left(x+2\right)\ge5x\)
\(\Leftrightarrow6x-9-15x+30\ge5x\)
\(\Leftrightarrow6x-15x-5x\ge9+30\)
\(\Leftrightarrow-14x\ge-21\)
\(\Leftrightarrow x\le\dfrac{21}{14}\le\dfrac{3}{2}\)
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0 3/2
lâu rồi cũng không nhớ cách làm :v
A
1
24 tháng 2 2022
\(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x-y\right)}{\left(x-y\right)^2x\left(x+y\right)}=\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{x\left(x-y\right)^2\left(x+y\right)}=\dfrac{x^2+y^2}{x}\)
H9
HT.Phong (9A5)
CTVHS
21 tháng 9 2023
\(9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3-x+y\right)\left(3+x-y\right)\)
____
\(\left(x-y\right)^2-4\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
____
\(\left(x+2\right)^2-y^2\)
\(=\left[\left(x+2\right)-y\right]\left[\left(x+2\right)+y\right]\)
\(=\left(x-y+2\right)\left(x+y+2\right)\)
____
\(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)
\(=2x\left(4x+2\right)\)
\(=4x\left(2x+1\right)\)
____
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=2x\cdot2y\)
\(=4xy\)
____
\(\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1-2x-y\right)\left(2xy+1+2x+y\right)\)
\(=\left[2x\left(y-1\right)-\left(y-1\right)\right]\left[2x\left(y+1\right)+\left(y+1\right)\right]\)
\(=\left(y-1\right)\left(2x-1\right)\left(2x+1\right)\left(y+1\right)\)
27 tháng 12 2021
a: Xét ΔABM và ΔADM có
AB=AD
\(\widehat{BAM}=\widehat{DAM}\)
AM chung
Do đó: ΔABM=ΔADM
14 tháng 12 2021
\(a,PT\left(1\right)=\dfrac{75y^4}{42x^2y^5};PT\left(2\right)=\dfrac{28x}{42x^2y^5}\\ b,PT\left(1\right)=\dfrac{11y^2}{102x^4y^3};PT\left(2\right)=\dfrac{9x^3}{10x^4y^3}\\ c,PT\left(1\right)=\dfrac{3x\left(3x+1\right)}{36x^2y^4};PT\left(2\right)=\dfrac{4y\left(y-2\right)}{36x^2y^4}\\ d,PT\left(1\right)=\dfrac{6y^2}{36x^3y^4};PT\left(2\right)=\dfrac{4x\left(x+1\right)}{36x^3y^4};PT\left(3\right)=\dfrac{9x^2y\left(x-1\right)}{36x^3y^4}\)
\(e,PT\left(1\right)=\dfrac{12y^4\left(3+2x\right)}{120x^4y^5};PT\left(2\right)=\dfrac{75x^2y^3}{120x^4y^5};PT\left(3\right)=\dfrac{8x^3}{120x^4y^5}\\ f,PT\left(1\right)=\dfrac{3\left(x+1\right)\left(4x-4\right)}{6x\left(x+3\right)\left(x+1\right)};PT\left(2\right)=\dfrac{2\left(x+3\right)\left(x-3\right)}{6x\left(x+1\right)\left(x+3\right)}\)
14 tháng 12 2021
\(g,PT\left(1\right)=\dfrac{4x^2}{2x\left(x+2\right)^3};PT\left(2\right)=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^3}\\ h,PT\left(1\right)=\dfrac{5}{3x\left(x-2\right)\left(x+2\right)}=\dfrac{10\left(x+3\right)}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\\ PT\left(2\right)=\dfrac{3}{2\left(x+2\right)\left(x+3\right)}=\dfrac{9x\left(x-2\right)}{6x\left(x+2\right)\left(x+3\right)\left(x-2\right)}\)
A
1
25 tháng 2 2022
\(=\dfrac{2x^2y^2}{3xy^2}-\dfrac{2ax+3x}{3a}=\dfrac{2x}{3}-\dfrac{2ax+3x}{3a}\)
\(=\dfrac{2xa-2xa-3x}{3a}=\dfrac{-3x}{3a}=-\dfrac{x}{a}\)
A
GIÚP MIK VỚI Ạ!! MIK ĐANG CẦN GẤP
1
25 tháng 2 2022
\(=\dfrac{5}{3}-\dfrac{5a-6}{3a}=\dfrac{5a-5a+6}{3a}=\dfrac{6}{3a}=\dfrac{2}{a}\)
a,Ta có:A=-4(x2-2xy+y2)-(y2-10y+25)+37
= -4(x-y)2-(y-5)2+37
Vì -4(x-y)2≤0 ∀x,y
- (y-5)2 ≤0 ∀y
⇒ A= -4(x-y)2-(y-5)2+37 ≤37
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-5=0\end{matrix}\right.\Leftrightarrow x=y=5\)
Vậy,Max A=37⇔x=y=5
b,B=-x2-y2+xy+2x+2y
⇔4B=-4x2-4y2+4xy+8x+8y
= -[4x2-4x(y-2)+(y2-4y+4)]-3(\(y^2-2.\dfrac{2}{3}y+\dfrac{4}{9}\))+\(\dfrac{16}{3}\)
\(=-\left(2x-y+2\right)^2-3\left(y-\dfrac{2}{3}\right)^2+\dfrac{16}{3}\)
Vì \(-\left(2x-y+2\right)^2\le0\forall x,y\)
\(-3\left(y-\dfrac{2}{3}\right)^2\le0\forall y\)
\(\Rightarrow4B=-\left(2x-y+2\right)^2-3\left(y-\dfrac{2}{3}\right)^2+\dfrac{16}{3}\le\dfrac{16}{3}\)
\(\Leftrightarrow B\le\dfrac{4}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x-y+2=0\\y-\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
Vậy,Max B=\(\dfrac{4}{3}\Leftrightarrow x=-\dfrac{2}{3};y=\dfrac{2}{3}\)
AMax chứ ko phải AMin