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bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc
1). x2y2(y-x)+y2z2(z-y)-z2x2(z-x)
2)xyz-(xy+yz+xz)+(x+y+z)-1
3)yz(y+z)+xz(z-x)-xy(x+y)
5)y(x-2z)2+8xyz+x(y-2z)2-2z(x+y)2
6)8x3(y+z)-y3(z+2x)-z3(2x-y)
7) (x2+y2)3+(z2-x2)3-(y2+z2)3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
(I): 4 x 2 + 4 x – 9 y 2 + 1 = ( 4 x 2 + 4 x + 1 ) – 9 y 2 = ( 2 x + 1 ) 2 – ( 3 y ) 2
= (2x + 1 + 3y)(2x + 1 – 3y) nên (I) đúng
Và
(II):
5 x 2 – 10 x y + 5 y 2 – 20 z 2 = 5 ( x 2 – 2 x y + y 2 – 4 z 2 ) = 5 [ ( x – y ) 2 – ( 2 z ) 2 ]
= 5(x – y – 2z)(x – y + 2z) nên (II) sai
Đáp án cần chọn là: A
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a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)
\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)
\(=\left(3x-5y\right)\left(2x-y\right)\)
b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)
\(=\left(x-5\right)^2-3^2\left(x+y\right)^2\)
\(=\left(x-5\right)^2-\left(3x+3y\right)^2\)
\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)
\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)
a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)
e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)
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\(\left\{{}\begin{matrix}3x-6y+2z=-4\\3x-y-3z=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x-6y+2z=-4\\3x-y-3z=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3x-6y=-4-2z\\3x-y=1+3z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5y=1+3z+4+2z\\3x-y=1+3z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5y=5+5z\\3x=y+1+3z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=1+z\\3x=1+z+1+3z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=1+z\\x=\dfrac{4z+6}{3}\end{matrix}\right.\)
\(S=9x^2-8\left(y^2+z^2\right)\)
\(S=9\left(\dfrac{4z+2}{3}\right)^2-8\left[\left(1+z\right)^2+z^2\right]\)
\(S=9.\dfrac{16z^2+16z+4}{9}-8\left[1+2z+z^2+z^2\right]\)
\(S=16z^2+16z+4-8-16z-16z^2\)
\(S=-4\)
Đính chính \(x=\dfrac{4z+2}{3}\) không phải \(x=\dfrac{4z+6}{3}\)
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a: Ta có: \(\left(x+y\right)^2\)
\(=x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)
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a: \(=4xy\left(1-5x^2y\right)\)
b: \(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
c: \(=x\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(x+y\right)\)
d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
** Lần sau bạn lưu ý ghi đề bài đầy đủ.
Cho $x,y,z$ là các số thực. CMR $x^2+y^2+z^2\geq xy+yz+xz$
----------------------------
Ta có:
BĐT cần cm tương đương với:
$x^2+y^2+z^2-xy-yz-xz\geq 0$
$\Leftrightarrow 2x^2+2y^2+2z^2-2xy-2yz-2xz\geq 0$
$\Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2xz+x^2)\geq 0$
$\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0$
(luôn đúng với mọi số thực $x,y,z$)
Do đó ta có đpcm
Dấu "=" xảy ra khi $x=y=z$
\(x^2+y^2+z^2=xy+3x+2z-4\)
\(\Leftrightarrow x^2+y^2+z^2-xy-3x-2z+4=0\)
\(\Leftrightarrow\left(\dfrac{1}{4}x^2-xy+y^2\right)+\left(z^2-2z+1\right)+3.\left(\dfrac{1}{4}x^2-x+1\right)=0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-y\right)^2+\left(z-1\right)^2+3.\left(\dfrac{1}{2}x-1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=y\\z=1\\\dfrac{1}{2}x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\z=1\\x=2\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(2;1;1\right)\)
=>x^2+y^2+z^2-xy-3y-2z+4=0
=>x^2-xy+1/4y^2+3/4y^2-3y+3+z^2-2z+1=0
=>(x-1/2y)^2+3/4(y-2)^2+(z-1)^2=0
=>(x-1/2y)^2=0 (y-2)^2=0 (z-1)^2=0
=>x=1/2y y=2 z=1
=>x=1,y=2,z=1