1 . Tính giá trị biểu thức
a) ( x + y + z )( x+ y + z)
b) ( x- y + z )(x- y - z )
c) ( x - 1 + y )(x - 1 - y )
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\(a,A=\left(x+y\right)^2-9z^2=\left(x+y-3z\right)\left(x+y+3z\right)\\ A=\left(5+7-36\right)\left(5+7+36\right)=-24\cdot48=-1152\\ b,B=\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)=\left(2x+y\right)\left(2x-y-1\right)\\ B=\left(2+2\right)\left(2-2-1\right)=4\cdot\left(-1\right)=-4\)
a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128
b: \(B=\left(2x-4\right)^2+2\cdot\left(2x-4\right)\left(x+1\right)+\left(x+1\right)^2\)
=(2x-4+x+1)^2
=(3x-3)^2
Khi x=-1/2 thì B=(-3/2-3)^2=(-9/2)^2=81/4
c: \(C=x^2\left(5-4\right)+y^2\left(4-6\right)+z^2\left(6+4\right)\)
=x^2-2y^2+10z^2
=6^2-2*5^2+10*4^2
=146
d: x=9 thì x+1=10
\(D=x^{2017}-x^{2016}\left(x+1\right)+x^{2015}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-\left(x+1\right)\)
=x^2017-x^2017+x^2016+...-x^3-x^2+x^2+x-x-1
=-1
a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128
Lời giải:
Ta có:
$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=2023.\frac{2024}{2023}$
$\Leftrightarrow 1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1=2024$
$\Leftrightarrow 3+\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}=2024$
$\Leftrightarrow 3+B=2024$
$\Leftrightarrow B=2021$
\(a,A=5x^2a-10xya+5y^2a\)
\(=5a\left(x^2-2xy+y^2\right)\)
\(=5a\left(x-y\right)^2\)
Thay x = 124; y=24;a=2 ta có
\(5.2\left(124-24\right)^2=10.100^2=100000\)
\(b,B=2x^2+2y^2-x^2z+z-y^2z-2\)
\(=2\left(x^2+y^2-1\right)-z\left(x^2+y^2-1\right)\)
\(=\left(x^2+y^2-1\right)\left(2-z\right)\)
Thay x = 1 ; y = 1; z= -1 ta có
\(\left(1^2+1^2-1\right)\left(2-\left(-1\right)\right)=\left(1+1-1\right)\left(2+1\right)=1.3=3\)
\(c,C=x^2-y^2+2y-1\)
\(=x^2-\left(y^2-2y+1\right)=x^2-\left(y-1\right)^2=\left(x-y+1\right)\left(x+y-1\right)\)
Thay x = 75; y = 26 ta có
\(\left(75-26+1\right)\left(75+26-1\right)=50.100=5000\)
Ta có: x-y-z = 0
\(\Rightarrow\) x = y+z
\(\Rightarrow\)y = x-z
\(\Rightarrow\)z = x-y
Thay vào B ta suy ra: \(\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
= \(\left(1-\frac{x-y}{x}\right)\left(1-\frac{y+z}{y}\right)\left(1+\frac{x-z}{z}\right)\)
= \(\left(\frac{-y}{x}\right).\left(\frac{z}{y}\right).\left(\frac{x}{z}\right)\)
= -y/y
= -1
Vậy B = -1
\(a.\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+xy+y^2+zy+zx+zy+z^2=x^2+y^2+z^2+2xy+2zy+2zx\)
\(b.\left(x-y+z\right)\left(x-y-z\right)=x^2-xy-zx-xy+y^2+zy+zx-zy-z^2=x^2+y^2-z^2-2xy\)
\(c.\left(x-1+y\right)\left(x-1-y\right)=x^2-x-xy-x+1+y+xy-y-y^2=x^2-y^2-2x+1\)
a) = \(^{\left(x+y+z\right)^2}\)=\(x^2\)+\(y^2\)+\(z^2\)+ 2xy +2xz+2yz
b) = \(\left(x-y\right)^2\)-\(z^2\)=\(x^2\)- 2xy+\(y^2\)-\(z^2\)
c)= \(\left(x-1\right)^2\)-\(y^2\)= \(x^2\)-2x+1 - \(y^2\)