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1A,B,D
2 M=2
3 \(=\dfrac{3}{4x}\)
4 \(=\dfrac{4\left(x+y\right)}{x-y}=\dfrac{4x+4y}{x-y}\)
5 K rút gọn đc
6 \(=\dfrac{4\left(x-1\right)+2\left(x-1\right)}{6\left(x-1\right)}=\dfrac{6\left(x-1\right)}{6\left(x-1\right)}=1\)
Ta có \(x^2-y^2-z^2=0\Rightarrow z^2=x^2-y^2\)
Có \(VT=\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-\left(4z\right)^2\)\(=\left(5x-3y\right)^2-16z^2=\left(5x-3y\right)^2-16\left(x^2-y^2\right)\)
\(=25x^2-30xy+9y^2-16x^2+16y^2=9x^2-30xy+25y^2\)
\(=\left(3x\right)^2-2.3x.5y+\left(5y\right)^2=\left(3x-5y\right)^2=VP\left(đpcm\right)\)
Lời giải:
a. $=(2x)^2-2.2x.5y+(5y)^2=4x^2-20xy+25y^2$
b. $=(3x)^2+2.3x.2y+(2y)^2=9x^2+12xy+4y^2$
c. $=(4y+3x)(4y-3x)=(4y)^2-(3x)^2=16y^2-9x^2$
\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)^2-\left(4z\right)^2=\left(3x-5y\right)^2\)
\(\Rightarrow\left(5x-3y\right)-16z^2-\left(3x-5y\right)^2=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-\left(9x^2-30xy+25y^2\right)=0\)
\(\Rightarrow25x^2-30xy+9y^2-16z^2-9x^2+30xy-25y^2=0\)
\(\Rightarrow25\left(x^2-y^2\right)+9\left(x^2-y^2\right)-16z^2=0\)
\(\Rightarrow34\left(x^2-y^2\right)-16z^2=0\)
a: \(3x^2y\left(2x^2-xy+5y^2\right)=6x^4y-3x^3y^2+15x^2y^3\)
b: \(\left(x+2\right)\left(x^2+3x-4\right)\)
\(=x^3+3x^2-4x+2x^2+6x-8\)
\(=x^3+5x^2+2x-8\)
\(\left(3x-2y\right)^3+\left(y+2x\right)^3-\left(4x-5y\right)\left(16x^2+20xy+25y^2\right)\)
\(=27x^3-54x^2y+36xy^2-8y^3+y^3+6xy^2+12x^2y+8x^3-\left(64x^3-125y^3\right)\)
\(=35x^3-42x^2y+42xy^2-7y^3-64x^3+125y^3\)
\(=-29x^3-42x^2y+42xy^2+118y^3\)
`a, (3x+1)^2 = 9x^2 + 6x + 1`.
`b, (4x+5y)^2 = 16x^2 + 20xy + 25y^2`
`c, (5x-1/2)^2 = 25x^2 - 5x + 1/4`
`d, (-x+2y^2)^2 = x^2 - 2y^2x + 4y^4`.
`a)`
`(3x + 1)^2`
`= 9x^2 + 6x + 1`
`b)`
`(4x + 5y)^2`
`= 16x^2 + 40xy + 25y^2`
`c)`
`(5x - 1/2)^2`
`= 25x^2 - 5x + 1/4`
`d)`
`(-x + 2y^2)^2`
`= x^2 - 4xy^2 + 4y^4`