Dúng hẳng đẳng thức để tính
a)(x+2y-3z-t)(x+2y+3z+t)
b)(2+x2+3x3+y2)(2-x2+3x3-y2)
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a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)
b: \(=-5y-9+xy\)
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
a) 3x(x + 1) - 5y(x + 1)
= (x + 1)(3x - 5y)
b) 3x(x - 6) - 2(x - 6)
= (x - 6)(3x - 2)
c) 4y(x - 1) - (1 - x)
= 4y(x - 1) + (x - 1)
= (x - 1)(4y + 1)
d) (x - 3)³ + 3 - x
= (x - 3)³ - (x - 3)
= (x - 3)[(x - 3)² - 1]
= (x - 3)(x - 3 - 1)(x - 3 + 1)
= (x - 3)(x - 4)(x - 2)
e) 7x(x - y) - (y - x)
= 7x(x - y) + (x - y)
= (x - y)(7x + 1)
h) 3x³(2y - 3z) - 15x(2y - 3z)²
= (2y - 3z)[3x³ - 15x(2y - 3x)]
= 3x(2y - 3x)[x² - 5(2y - 3x)]
= 3x(2y - 3x)(x² - 10y + 3x)
= 3x(2y - 3x)(x² + 3x - 10y)
k) 3x(x + 2) + 5(-x - 2)
= 3x(x + 2) - 5(x + 2)
= (x + 2)(3x - 5)
l) 18x²(3 + x) + 3(x + 3)
= (x + 3)(18x² + 3)
= 3(x + 3)(6x² + 1)
m) 7x(x - y) - (y - x)
= 7x(x - y) + (x - y)
= (x - y)(7x + 1)
n) 10x(x - y) - 8y(y - x)
= 10x(x - y) + 8y(x - y)
= (x - y)(10x + 8y)
= 2(x - y)(5x + 4y)
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
a) 2x.(x2 - 7x - 3)
= 2xx2 + 2x(-7x) + 2x(-3)
= 2x2x - 2.7xx - 2.3x
= 2x3 - 14x2 - 6x
a) \(2x\left(x^2-7x-3\right)=2x.x^2-2x.7x-2x.3=2x^3-14x^2-6x\)
b) \(\left(-2x^3+y^2-7xy\right)4xy^2=\left(-2x^3\right)4xy^2+y^24xy^2-7xy.4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)
c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-5x^32x^2-5x^33x-5x^3.-5=-10x^5-15x^4+25x^3\)
d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-3x^32x^2-3x^3.-xy-3x^3y^2=-6x^5+3x^4y-3x^3y^2\)
e) \(\left(x^2-2x+3\right)\left(x-4\right)=x\left(x^2-2x+3\right)-4\left(x^2-2x+3\right)=x^3-2x^2+3x-4x^2+8x-12=x^3-6x^2+11x-12\)
f) \(\left(2x^3-3x-1\right)\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)
2,
M + N = 3xyz - 3x2 + 5xy - 1 + 5x2 + xyz - 5xy + 3 - y
= -3x2 + 5x2 + 3xyz + xyz + 5xy - 5xy - y - 1 + 3
= 2x2 + 4xyz - y +2.
M - N = (3xyz - 3x2 + 5xy - 1) - (5x2 + xyz - 5xy + 3 - y)
= 3xyz - 3x2 + 5xy - 1 - 5x2 - xyz + 5xy - 3 + y
= -3x2 - 5x2 + 3xyz - xyz + 5xy + 5xy + y - 1 - 3
= -8x2 + 2xyz + 10xy + y - 4.
N - M = (5x2 + xyz - 5xy + 3 - y) - (3xyz - 3x2 + 5xy - 1)
= 5x2 + xyz - 5xy + 3 - y - 3xyz + 3x2 - 5xy + 1
= 5x2 + 3x2 + xyz - 3xyz - 5xy - 5xy - y + 3 + 1
= 8x2 - 2xyz - 10xy - y + 4.
3,
a) P + (x2 – 2y2) = x2 – y2 + 3y2 – 1
P = (x2 – y2 + 3y2 – 1) - (x2 – 2y2)
P = x2 – y2 + 3y2 – 1 - x2 + 2y2
P = x2 – x2 – y2 + 3y2 + 2y2 – 1
P = 4y2 – 1.
Vậy P = 4y2 – 1.
b) Q – (5x2 – xyz) = xy + 2x2 – 3xyz + 5
Q = (xy + 2x2 – 3xyz + 5) + (5x2 – xyz)
Q = xy + 2x2 – 3xyz + 5 + 5x2 – xyz
Q = 7x2 – 4xyz + xy + 5
Vậy Q = 7x2 – 4xyz + xy + 5.
4,
a, Thu gọn : x2+2xy-3x3+2y3+3x3-y3
= x2+2xy+(-3x3+3x3)+2y3-y3
=x2+2xy+2y3-y3
Thay x=5,y=4 vào đa thức x2+2xy+2y3-y3 Ta có:
52 + 2.5.4 + 43 = 25 + 40 + 64 = 129.
Vậy giá trị của đa thức x2+2xy+2y3-y3 tại x=5,y=4 là 129
b,
Thay x = -1; y = -1 vào biểu thức xy-x2y2+x4y4-x6y6+x8y8 Ta Có
M = (-1)(-1) - (-1)2.(-1)2 + (-1)4. (-1)4-(-1)6.(-1)6 + (-1)8.(-1)8
= 1 -1 + 1 - 1+ 1 = 1.
Vậy giá trị của biểu thức xy-x2y2+x4y4-x6y6+x8y8 tại x=-1, y=-1 là 1
5,
a, C=A+B
C = x2 – 2y + xy + 1 + x2 + y - x2y2 - 1
C = 2x2 – y + xy - x2y2
b) C + A = B => C = B - A
C = (x2 + y - x2y2 - 1) - (x2 – 2y + xy + 1)
C = x2 + y - x2y2 - 1 - x2 + 2y - xy - 1
C = - x2y2 - xy + 3y - 2.
\(x^2+10x+y^2-2y+26+\left(3z-6\right)^2=0\)
\(\Leftrightarrow x^2+10x+25+y^2-2y+1+\left(3z-6\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)^2+\left(y-1\right)^2+\left(3z-6\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x+5=0\\y-1=0\\3z-6=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=1\\z=2\end{cases}}\)
a)(x+2y-3z-t)(x+2y+3z+t)
=\(\left(x+2y\right)^2-\left(3z+t\right)^2\)