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a: \(F=-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3\)
\(=-\left(2x-y\right)^2\cdot\left[2x-y+x\right]-y^3\)
\(=-\left(2x-y\right)^2\cdot\left(3x-y\right)-y^3\)
\(=\left(-4x^2+4xy-y^2\right)\left(3x-y\right)-y^3\)
\(=-12x^3+4x^2y+12x^2y-4xy^2-3xy^2+y^3-y^3\)
\(=-12x^3+16x^2y-7xy^2\)
\(\left(x-2\right)^2+y^2=0\)
mà \(\left(x-2\right)^2+y^2>=0\forall x,y\)
nên dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\y=0\end{matrix}\right.\)
=>x=2 và y=0
Thay x=2 và y=0 vào F, ta được:
\(F=-12\cdot2^3+16\cdot2^2\cdot0-7\cdot2\cdot0^2\)
\(=-12\cdot2^3\)
\(=-12\cdot8=-96\)
b: \(G=\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=x^3+y^3+3\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)
\(=x^3+y^3+3\left(8x^3-y^3\right)\)
\(=x^3+y^3+24x^3-3y^3\)
\(=25x^3-2y^3\)
Ta có: \(\left\{{}\begin{matrix}x+y=2\\y=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-3\\x=2-y=2-\left(-3\right)=2+3=5\end{matrix}\right.\)
Thay x=5 và y=-3 vào G, ta được:
\(G=25\cdot5^3-2\cdot\left(-3\right)^3\)
\(=25\cdot125-2\cdot\left(-27\right)\)
\(=3125+54=3179\)
c: \(H=\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)
\(=x^3+27y^3+27x^3-y^3\)
\(=28x^3-26y^3\)
Ta có: \(\left\{{}\begin{matrix}3x-y=5\\x=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=3x-5=3\cdot2-5=1\end{matrix}\right.\)
Thay x=2 và y=1 vào H, ta được:
\(H=28\cdot2^3-26\cdot1^3\)
\(=28\cdot8-26\)
=198
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a: M=2(-2x-3xy^2+1)-3xy^2+1
=-4x-6xy^2+2-3xy^2+1
=-4x-9xy^2+3
b: Thay x=-2 và y=3 vào M, ta được:
M=2*(-2)-3*(-2)*3^2+1
=-4+1+6*9
=54-3
=51
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Ta có:
\(2x+y=11z\) và \(3x-y=4z\)
Chia theo vế ta có:
\(\dfrac{2x+y}{3x-y}=\dfrac{11z}{4z}=\dfrac{11}{4}\)
\(\Leftrightarrow4\left(2x+y\right)=11\left(3x-y\right)\)
\(\Leftrightarrow8x+4y=33x-11y\)
\(\Leftrightarrow15y=25x\)
\(\Leftrightarrow3y=5x\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}=k\)
\(\Rightarrow x=3k,y=5k\)
Thay vào Q ta có:
\(Q=\dfrac{2\cdot\left(3k\right)^2-3\cdot3k\cdot5k}{\left(3k\right)^2+3\cdot\left(5y\right)^2}\)
\(Q=\dfrac{18k^2-45k^2}{9k^2+75k^2}\)
\(Q=\dfrac{k^2\left(18-45\right)}{k^2\left(9+75\right)}\)
\(Q=\dfrac{-27}{84}=-\dfrac{9}{28}\)
\(\dfrac{2x+y}{3x-y}=\dfrac{11}{4}\)
=>33x-11y=8x+4y
=>25x=15y
=>5x=3y
=>x/3=y/5=k
=>x=3k; y=5k
\(Q=\dfrac{2\cdot9k^2-3\cdot3k\cdot5k}{9k^2+3\cdot25k^2}=\dfrac{18-9\cdot5}{9+3\cdot25}=\dfrac{-9}{28}\)
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\(A=\frac{2x-y}{3x-y}+\frac{5y-x}{3x+y}\)
\(=\frac{\left(2x-y\right)\left(3x+y\right)+\left(5y-x\right)\left(3x-y\right)}{\left(3x-y\right)\left(3x+y\right)}\)
\(=\frac{3x^2+15xy-6y^2}{9x^2-y^2}\)
\(=\frac{3\left(x^2+5xy-2y^2\right)}{9x^2-y^2}\)
\(=\frac{3\left(10x^2+5xy-3y^2-9x^2+y^2\right)}{9x^2-y^2}\)
\(=-\frac{3\left(9x^2-y^2\right)}{9x^2-y^2}\)
= - 3 (đpcm)
~~~
\(A=\frac{1}{x}+\frac{1}{x+2}+\frac{x-2}{x^2+2x}\)
\(=\frac{x+2+x+x-2}{x^2+2x}\)
\(=\frac{3x}{x\left(x+2\right)}\)
\(=\frac{3}{x+2}\)
\(A\in Z\)
\(\Leftrightarrow3⋮x+2\)
\(\Leftrightarrow x+2\in\text{Ư}\left(3\right)=\left\{-3:-1;1;3\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-1;1\right\}\)
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a: A=2/3x^2y+4x^2y=14/3x^2y
=14/3*9*7=294
b: B=xy^2(1/2+1/3+1/6)=xy^2=3/4*1/4=3/16
c: C=x^3y^3(2+10-20)=-8x^3y^3
=-8*1^3(-1)^3=8
d: D=xy^2(2018+16-2016)
=18xy^2
=18(-2)*1/9=-4
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Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
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Bài 4:
Ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
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Vì pt ẩn x của f(x,y) = 0 nhận x=2 làm nghiệm nên ta có:
\(\left(3.2-5y+4\right)\left(2.2+3y-2\right)=0\\ \Leftrightarrow\left(10-5y\right)\left(2+3y\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}10-5y=0\\2+3y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=2\\y=\dfrac{-2}{3}\end{matrix}\right.\)
Vậy để pt ẩn x của f(x,y) = 0 nhận x=2 làm nghiệm thì \(y=2\) hoặc \(y=\dfrac{-2}{3}\)
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\(A=4x^2+12xy+9y^2\)
\(B=25x^2-10xy+y^2\)
\(C=8x^3+12x^2y^2+6xy^4+y^6\)
\(D=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4y^2}{25}\)
\(E=x^3-27y^3\)
\(F=x^6-27\)