Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(x^2-y^2-z^2=0\Rightarrow z^2=x^2-y^2\)
Có \(VT=\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-\left(4z\right)^2\)\(=\left(5x-3y\right)^2-16z^2=\left(5x-3y\right)^2-16\left(x^2-y^2\right)\)
\(=25x^2-30xy+9y^2-16x^2+16y^2=9x^2-30xy+25y^2\)
\(=\left(3x\right)^2-2.3x.5y+\left(5y\right)^2=\left(3x-5y\right)^2=VP\left(đpcm\right)\)
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)-16z^2-\left(3x-5y\right)^2=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-\left(9x^2-30xy+25y^2\right)=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-9x^2+30xy-25y^2=0\)
\(\Rightarrow25\left(x^2-y^2\right)+9\left(x^2-y^2\right)-16z^2=0\)
\(\Rightarrow34\left(x^2-y^2\right)-16z^2=0\)
a) Đề sai nha bạn :) mấy dấu cộng bạn phỉa chuyển thành dấu nhân nhé
\(A=\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)
\(A=2^{512}-1+1\)
\(A=2^{512}\)
b . ( 5x - 3y + 4z )( 5x - 3y - 4z ) = ( 5x - 3y )^2 - ( 4z )^2 = 25x^2 - 30xy + 9y^2 - 16z^2 = 25( y^2 + z^2 ) - 30xy + 9y^2 - 16z^2 = 9z^2 + 34y^2 - 30xy ( 1 )
( 3x - 5y )^2 = 9x^2 - 30xy + 25y^2 = 9( y^2 + z^2 ) - 30xy + 25y^2 = 34y^2 + 9z^2 - 30xy ( 2 )
Tu ( 1 ) va ( 2 ) => dpcm
4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm
\(2x^2-4x=2x\left(x-2\right)\)
\(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
\(10\left(x-y\right)-6x\left(y-x\right)=10\left(x-y\right)+6x\left(x-y\right)=\left(10+6x\right)\left(x-y\right)=2\left(x-y\right)\left(3x+5\right)\)\(\left(x+1\right)^2-25=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
\(x^2+3x-y^2+3y=\left(x-y\right)\left(x+y\right)+3\left(x+y\right)=\left(x+y\right)\left(x-y+3\right)\)
\(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\)
\(x^2-7x-y^2+7y=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\)
\(3y^2-3z^2+3x^2=3\left(y^2-z^2+x^2\right)\)
Ta có 2x=3y =>\(\dfrac{x}{3}=\dfrac{y}{2}\)=>\(\dfrac{x}{12}=\dfrac{y}{8}\)
5y=4z=>\(\dfrac{y}{4}=\dfrac{z}{5}\)=>\(\dfrac{y}{8}=\dfrac{z}{10}\)
=>\(\dfrac{x}{12}=\dfrac{y}{8}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được
\(\dfrac{x^2}{144}=\dfrac{y^2}{64}=\dfrac{z^2}{100}=\)\(\dfrac{x^2}{144}=\dfrac{3y^2}{192}=\dfrac{2z^2}{200}\)
=\(\dfrac{342}{152}=\dfrac{9}{4}\)
Còn lại tự làm nha
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
\(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)
\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow2\left(x-1\right)^2+3\left(x+y\right)^2+\left(z+2\right)^2=0\)
Vì \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\\3\left(x+y\right)^2\ge0\\\left(z+2\right)^2\ge0\end{matrix}\right.\)\(\forall x;y;z\) Nên \(2\left(x-1\right)^2+3\left(x+y\right)^2+\left(z+2\right)^2\ge0\forall x;y;z\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)^2=0\\3\left(x+y\right)^2=0\\\left(z+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=-2\end{matrix}\right.\)
Vậy \(x=1;y=-1;z=-2\)
Ta có:
\(x^2-y^2-z^2=0\)
\(16x^2-16y^2-16z^2=0\)
\(25x^2-9x^2+9y^2-25y^2-16z^2+30xy-30xy=0\)
\(\left(5x-3y\right)^2-16z^2= \left(3x-5y\right)^2\)
\(\left(5x-3y-4z\right)\left(5x-3y+4z\right)=\left(3x-5y\right)^2\)