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Bài 1 bạn tách hằng đẳng thức ra rồi thay vào tính bình thường . Mình làm bài 2 nha.
D = ( x + y )2 - 6.( x + y ) - 5
Thay x + y = -9 vào D, ta có :
D = ( -9 )2 - 6.( -9 ) - 5 = 81 + 54 - 5 = 130
Bài 1:
a: \(=\left(3x-2\right)^2+2\left(3x-2\right)\left(3x+2\right)+\left(3x+2\right)^2\)
\(=\left(3x-2+3x+2\right)^2=36x^2=36\cdot\dfrac{1}{9}=4\)
b: \(=\left(x+y-7-y+6\right)^2=\left(x-1\right)^2=100^2=10^4\)
c: \(C=4x^2-20x+27\)
\(=4x^2-20x+25+2\)
\(=\left(2x-5\right)^2+2\)
\(=\left(2\cdot52.5-5\right)^2+2=100^2+2=10002\)
a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)
b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)
\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)
c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)
d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)
e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)
= 31
f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)
a) \(\left(x-10\right)^2-x\left(x+8\right)=-12x+100=-11,76+100=88,24\)
b) \(x^3-9x^2+27x-27=\left(x-3\right)^3=\left(5-3\right)^3=8\)
c) \(6x\left(2x-7\right)-\left(3x-5\right)\left(4x+7\right)=-43x+35=121\)
\(a)\) \(\left(x-10\right)^{^2}-x.\left(x+8\right)\) \(với\) \(x=0,98\)
\(=-12x+100\)
\(=-11,76+100\)
\(=88,24\)
\(b)\) \(x^3-9x^2+27.x-27\) \(với\) \(x=5\)
\(=\left(x-3\right)^3\)
\(=\left(5-3\right)^3\)
\(=8\)
\(c)\)\(6x.\left(2x-7\right)-\left(3x-5\right).\left(4x+7\right)\) \(tại\) \(x=-2\)
\(=-43+35\)
\(=121\)
Chúc bạn hôc tốt nha ❤
a. \(=4x^2-4xy+y^2+4x^2-4xy+y^2=8x^2+2y^2\)
\(=8.\left(\frac{1}{21}\right)^2+4.\left(-0.3\right)^2=\frac{4169}{11025}\)
b, \(=\left(\frac{1}{7}xy+7yz+\frac{1}{7}xy-7yz\right)\left(\frac{1}{7}xy+7yz-\frac{1}{7}xy+7yz\right)\)
\(=\frac{2}{7}xy.14yz=4xy^2z=4.2.\left(0,25\right)^2.\left(-4\right)=-2\)
b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)
\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)
\(=-x^2+18xy\)
c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
\(=\left(2a-3b\right)^2-16c^2\)
\(=4a^2-12ab+9b^2-16c^2\)
B) \(\left|3x-6\right|+\left|x+1\right|=x\)
\(\Leftrightarrow\left|3x-6\right|+\left|x+1\right|-x=0\)
TH1: \(x\le-1\)
6-3x-x-1-x=0\(\Leftrightarrow\) x=\(-1\) (thỏa mãn)
TH2: \(-1< x\le2\)
6-3x +x+1-x=0\(\Leftrightarrow\) x= \(\dfrac{7}{3}\)( thỏa mãn)
TH3 : x>2
3x-6+x+1-x=0\(\Leftrightarrow x=\dfrac{5}{3}\) (loại)
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: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)
\(=\left(x+y-7-y+6\right)^2\)
\(=\left(x-1\right)^2=100^2=10000\)
`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`
`= 2xy`.
Thay `x = 2/3; y = -3/4` vào BT:
`2 . 2/3 . -3/4 = -1.`
`b, x(x-2y) - y(y^2-2x)`
`= x^2 - 2xy - y^3 + 2xy`
`= x^2 - y^3`
Thay `x = 5; y =3` vào BT:
`= 5^2 - 3^3 = 25 - 27 = -2`
a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)
\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)
\(=2xy\)
Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:
\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)
b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)
\(=x^2-2xy-y^3+2xy\)
\(=x^2-y^3\)
Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)
a, \(A=\left(3x-2\right)^2+\left(3x+2\right)^2+2\left(9x^2-4\right)\)
\(=\left(3x-2\right)^2+\left(3x+2\right)^2+2\left(3x-2\right)\left(3x+2\right)\)
\(=\left(3x-2+3x+2\right)^2\)
\(=36x^2=36.\left(-\frac{1}{3}\right)^2=4\)
b, \(B=\left(x+y-7\right)^2-2\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)
\(=\left[\left(x+y-7\right)-\left(y-6\right)\right]^2\)
\(=\left(x-1\right)^2\)
\(=\left(101-1\right)^2=10000\)
c, \(C=4x^2-20x+27\)
\(=\left(2x\right)^2-2.2x.5+5^2+2\)
\(=\left(2x-5\right)^2+2\)
\(=\left(52,5.2-5\right)^2+2\)
\(=100^2+2=10002\)
Bài này dễ mà chỉ dùng hằng đẳng thức thôi. Chúc bạn học tốt.