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a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(A=9x\)
Thay x = 15 vào, ta có:
\(A=9.15=135\)
b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(B=5x^2-20xy-4y^2+20xy\)
\(B=5x^2-4y\)
Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có:
\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)
c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)
\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(C=9x^2y^2-xy^3-8x^3\)
Thay \(x=\frac{1}{2};y=2\) vào, ta có:
\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)
d) \(D=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)
\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)
\(D=18x^2+12x-7\)
Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)
+) Với x = -2
\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)
+) Với x = 2
\(D=18.2^2+12.2-7=89\)
Ta có: \(A=\left(x-3\right)^2+\left(11-x\right)^2\)
\(=x^2-6x+9+x^2-22x+121\)
\(=2x^2-28x+130\)
\(=2\left(x^2-14x+49+16\right)\)
\(=2\left(x-7\right)^2+32\ge32\forall x\)
Dấu '=' xảy ra khi x=7
\(5x2+5y2+8xy-2x+2y+2=0\)
(=) \((4x^2 + 8xy + 4y^2) + (x^2 - 2x +1) + (y^2 + 2y +1) = 0 \)
(=) \(4(x+y)^2 + (x-1)^2 + (y+1)^2 = 0 \)
Ta có \(\begin{cases} 4(x+y)^2 ≥ 0 \\ (x-1)^2 ≥ 0 \\ (y+1)^2 ≥ 0 \end{cases} \)
=> \(4(x+y)^2 + (x-1)^2 + (y+1)^2 ≥ 0 \)
Vậy để \(4(x+y)^2 + (x-1)^2 + (y+1)^2 = 0 \)
(=) \(\begin{cases} 4(x+y)^2 = 0 \\ (x-1)^2 = 0 \\ (y+1)^2 = 0 \end{cases} \)
(=) \(\begin{cases} x = -y \\ x = 1 \\ y = -1 \end{cases} \)
(=) \(\begin{cases} x = 1 \\ y = -1 \end{cases} \)
Vậy \(M=(x+y)^{2015}+(x-2)^{2016}+(y+1)^{2017} M=(1-1)^{2015} + (1-2)^{2016} + (-1+1)^{2017} M=0^{2015} + (-1)^{2016} +0^{2017} M= 1 \)Vậy M = 1
a. Ta có : (x + y)[(x - y)2 + xy]
= (x + y)(x2 - 2xy + y2 + xy)
= (x + y)(x2 - xy + y2)
= x3 + y3
b. Ta có : x3 + y3 - xy(x + y)
= x3 + y3 - x2y - xy2
=x2(x - y) + y2(y - x)
= (x - y)(x2 - y2)
= (x - y)2.(x + y) đpcm
c) Ta có (x + y)3 - 3xy(x + y)
= (x + y)[(x + y)2 - 3xy)
= (x + y)(x2 + 2xy + y2 - 3xy)
= (x + y)(x2 - xy + y2) (đpcm)
a) VP = ( x + y )( x2 - 2xy + y2 + xy ) = ( x + y )( x2 - xy + y2 ) = x3 + y3 = VT ( đpcm )
b) VP = ( x + y )( x - y )2 = ( x + y )( x2 - 2xy + y2 ) = x3 - 2x2y + xy2 + x2y - 2xy2 + y3 = x3 + y3 - x2y - xy2 = x3 + y3 - xy( x + y ) = VT ( đpcm )
c) VP = x3 + 3x2y + 3xy2 + y3 - 3x2y - 3xy2 = x3 + y3 = ( x + y )( x2 - xy + y2 ) = VT ( đpcm )
Tính giá trị của $x+y-2=0$ là sao nhỉ? $x+y-2=0$ sẵn rồi mà bạn?
\(Q=x^2+y^2+xy+x+y+10\)
\(=\left(x^2+xy+x\right)+y^2+y+10\)
\(=x^2+x\left(y+1\right)+y^2+y+10\)
\(=x^2+2.x.\frac{y+1}{2}+\left(\frac{y+1}{2}\right)^2+y^2+y-\left(\frac{y+1}{2}\right)^2+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{\left(y+1\right)^2}{4}+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{y^2+2y+1}{4}+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+y^2+y-\frac{1}{4}y^2-\frac{1}{2}y-\frac{1}{4}+10\)
\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}y^2+\frac{1}{2}y+\frac{39}{4}\)
\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y^2+\frac{2}{3}y+13\right)=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left(y^2+2.y.\frac{2}{6}+\frac{4}{36}-\frac{4}{36}+13\right)\)
\(=\left(x+\frac{y+1}{2}\right)^2+\frac{3}{4}\left[\left(y+\frac{2}{6}\right)^2+\frac{116}{9}\right]=\left(\frac{2x+y+1}{2}\right)^2+\frac{3}{4}\left(y+\frac{2}{6}\right)^2+\frac{29}{3}\)
Vì \(\left(\frac{2x+y+1}{2}\right)^2\ge0;\frac{3}{4}\left(y+\frac{2}{6}\right)^2\ge0=>\left(\frac{2x+y+1}{2}\right)^2+\frac{3}{4}\left(y+\frac{2}{6}\right)^2+\frac{29}{3}\ge\frac{29}{3}>0\) (với mọi x;y)
Vậy biểu thức Q luôn dương với mọi giá trị của biến
=>4Q=4x2+4xy+4y2+4x+4y+40
=4x2+4x(y+1)+(y+1)2+4y2-y2+4y-2y+40-1
=(2x+y+1)2+3y2+2y+39
\(=\left(2x+y+1\right)^2+\left(\sqrt{3}y+\frac{\sqrt{3}}{3}\right)^2+\frac{116}{3}\)
\(\Rightarrow Q=\left(\frac{2x+y+1}{2}\right)^2+\left(\frac{\sqrt{3}y+\frac{\sqrt{3}}{3}}{2}\right)^2+\frac{29}{3}>0\)
=>đpcm
