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a) M = (x² + 3xy - 3x³) + (2y³ - xy + 3x³)
= x² + 3xy - 3x³ + 2y³ - xy + 3x³
= x² + (3xy - xy) + (-3x³ + 3x³) + 2y³
= x² + 2xy + 2y³
Tại x = 5 và y = 4
M = 5² + 2.5.4 + 2.4³
= 25 + 40 + 2.64
= 65 + 128
= 193
b) N = x²(x + y) - y(x² - y²)
= x³ + x²y - x²y + y³
= x³ + (x²y - x²y) + y³
= x³ + y³
Tại x = -6 và y = 8
N = (-6)³ + 8³
= -216 + 512
= 296
c) P = x² + 1/2 x + 1/16
= (x + 1/2)²
Tại x = 3/4 ta có:
P = (3/4 + 1/2)² = (5/4)² = 25/16
a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)
\(=-4.\dfrac{1}{4}+10=9\)
b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)
\(=\left(-2\right).\left(32-32\right)=0\)
a: Ta có: \(A=\left(2x-1\right)^2+\left(3-2x\right)\left(3+2x\right)\)
\(=4x^2-4x+1+9-4x^2\)
\(=-4x+10\)
\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)
\(a)\)
\(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(4x-6\right)+xy\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-3\right)+\left(2x-3\right)^2+xy\)
\(=\left(2x+3-2x+3\right)^2+xy\)
\(=6^2+2\left(-1\right)\)
\(=36-2\)
\(=34\)
\(b)\)
\(\left(x-2\right)^2-\left(x-1\right)\left(x+1\right)-x\left(1-x\right)\)
\(=x^2-4x+4-x^2+1-x+x^2\)
\(=x^2-5x+5\)
Thay \(x=-2\)vào ta có:
\(\left(-2\right)^2-5\left(-2\right)+5\)
\(=4+10+5\)
\(=19\)
Bài 2:
a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
hay \(x=\dfrac{2}{7}\)
b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Bài 1:
a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)
\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)
\(=xy\)
=1
b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)
\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)
\(=x^2-y^2\)
\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
a) Ta có: \(\left(3x-2\right)^2+2\left(3x-2\right)\left(3x+2\right)+\left(3x+2\right)^2\)
\(=\left(3x-2+3x+2\right)^2\)
\(=36x^2\)(1)
Thay \(x=-\dfrac{1}{3}\) vào biểu thức (1), ta được:
\(36\cdot\left(-\dfrac{1}{3}\right)^2=36\cdot\dfrac{1}{9}=4\)
b) Sửa đề: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)
Ta có: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)
\(=\left(x+y-7-y+6\right)^2\)
\(=\left(x-1\right)^2=100^2=10000\)
\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)
\(=2x^2-2xy-y^2+2xy\)
\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)
\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(=5x^2-20xy-4y^2+20xy\)
\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)
\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)
\(=-xy\left(x+1\right)\)
mình biết câu b rồi nhưng câu a thì chưa!
b) x^3(x+y)-x^2(x^2+xy)-x(x-y)
=x^4+x^3y-x^4-x^3y-x^2+xy
=-x^2+xy tại x=10,y=-5 ta có;
=-10^2+10(-5)
= 50
a: \(A=x^2-2xy+y^2+x^2+2xy+y^2-2x^2-x\)
=-x
=-2