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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) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)
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. \(=4x^2-4xy+y^2+4x^2-4xy+y^2=8x^2+2y^2\)
\(=8.\left(\frac{1}{21}\right)^2+4.\left(-0.3\right)^2=\frac{4169}{11025}\)
b, \(=\left(\frac{1}{7}xy+7yz+\frac{1}{7}xy-7yz\right)\left(\frac{1}{7}xy+7yz-\frac{1}{7}xy+7yz\right)\)
\(=\frac{2}{7}xy.14yz=4xy^2z=4.2.\left(0,25\right)^2.\left(-4\right)=-2\)
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
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.