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a) ĐKXĐ : \(x+y\ne0\)
\(x^2-2y^2=xy\)
\(x^2-y^2-y^2-xy=0\)
\(\left(x-y\right)\left(x+y\right)-y\left(y+x\right)=0\)
\(\left(x+y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\left(Loai\right)\\x-2y=0\left(Chon\right)\end{matrix}\right.\)
Với x - 2y = 0 ta có x = 2y
Thay x = 2y vào A ta có :
\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
\(x^2-xy-12y^2=0\)
\(\Leftrightarrow\left(x^2+3xy\right)-\left(4xy-12y^2\right)=0\)
\(\Leftrightarrow x\left(x+3y\right)-4y\left(x+3y\right)=0\)
\(\Leftrightarrow\left(x+3y\right)\left(x-4y\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3y\\x=4y\end{cases}}\)
TH1:\(x=-3y\)
\(A=\frac{3\cdot\left(-3y\right)+2y}{3\left(-3y\right)-2y}=\frac{-9y+2y}{-9y-2y}=\frac{-7y}{-11y}=\frac{7}{11}\)
TH2:\(x=4y\)
\(A=\frac{3\cdot4y+2y}{3\cdot4y-2y}=\frac{12y+2y}{12y-2y}=\frac{14y}{10y}=\frac{7}{5}\)
1
a
3x(x^2-3x+5)
= 3x^3- 9 x^2+15x
b
(3x+2y)(3x-2y)
= (3x)^2- (2y)^2
=9 x^2- 4 y^2
c
4x^2+4x+1:(2x+1)
= (2x+1)^2:(2x+1)
= (2x+1)
1. a) Ta có: \(x^2-2y^2=xy\) \(\Leftrightarrow\) \(x^2-xy-2y^2=0\)
\(\Leftrightarrow\) \(x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow\) \(x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\) \(\left(x+y\right)\left(x-2y\right)=0\)
Vì \(\left(x+y\right)\ne0\) nên \(x-2y=0\) hay \(x=2y\). Thay \(x=2y\) vào A, ta được:
\(A=\dfrac{\left(2y\right)^2-y^2}{\left(2y\right)^2+y^2}=\dfrac{4y^2-y^2}{4y^2+y^2}=\dfrac{3y^2}{5y^2}=\dfrac{3}{5}\)
Bài 1
a)\(=x^2+2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+2\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
MIN = \(-\frac{1}{4}\)khi \(x+\frac{3}{2}=0\Rightarrow x=-\frac{3}{2}\)
g. \(x^{^3}+3x^2+3x+1-27z^3\\ =\left(x^{^3}+3x^2+3x+1\right)-27z^3\\ =\left(x+1\right)^3-27z^3\\ =\left(x+1-3\right)\left[\left(x+1\right)^2+\left(x+1\right)3z+9z^2\right]\\ =\left(x-2\right)\left(x+2x+1+3zx+3z+9z^2\right)\\ =\left(x-2\right)\left(3x+3zx+3z+9z^2+1\right)\left(x-2\right)3x\left(1+z\right)+3z\left(1+z\right)+1\\ =\left(x-2\right)\left(1+z\right)\left(3x+3z\right)+1\\ =\left(x-2\right)\left(1-z\right)3\left(x+z\right)+1\)
Mk lm hơi tắt, bn chú ý nha:
a,\(x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
=\(\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
=\(\left(x+1\right)^2\left(x^2-x+1\right)\)
b,\(\left(x^4-x^3\right)-\left(x^2-1\right)\)
=\(x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
= \(\left(x-1\right)\left(x^3-x-1\right)\)
c,Đề phải thế này nha:
\(x^2y-xy^2-x+y\)=\(xy\left(x-y\right)-\left(x-y\right)\)
=\(\left(x-y\right)\left(xy-1\right)\)
d,hình như đề sai đó bn, thế này đúng ko?
\(a^2x+a^2y-7x-7y\)=\(a^2\left(x+y\right)-7\left(x+y\right)\)=\(\left(x+y\right)\left(a^2-7\right)\)
e,\(4x^2-x^2-16y^2+4y^2\)
=\((4x^2-16y^2)-\left(x^2-4y^2\right)\)
=\(4\left(x-2y\right)\left(x+2y\right)-\left(x^2-2y\right)\left(x^2+2y\right)\)=\(3\left(x-2y\right)\left(x+2y\right)\)
Cách này nhanh hơn:\(3\left(x^2-4y^2\right)\)
=\(3\left(x-2y\right)\left(x+2y\right)\)
g,\(\left(x+1\right)^3-\left(3z\right)^3\)=
\(\left(x-3z+1\right)[\left(x+1\right)^2+3z\left(x+1\right)+9z^2]\)Nếu thấy đề bn đưa sai thì nhắc mk nhé?
Mong các bn giúp đỡ thêm
Chúc các bn hc tốt
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)