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a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
1)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)
d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)
e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)
f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)
\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)
2)
a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )
d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
a) \(-2x\left(10x-3\right)+5x\left(4x+1\right)=25\)
\(-20x^2+6x+20x^2+5x=25\)
\(\Rightarrow6x+5x=25\)
\(\Rightarrow11x=25\)
\(\Rightarrow x=\dfrac{25}{11}\)
b) \(y\left(5-2y\right)+2y\left(y-1\right)=15\)
\(5y-2y^2+2y^2-2y=15\)
\(\Rightarrow5y-2y=15\)
\(\Rightarrow3y=15\)
\(\Rightarrow y=5\)
c)\(x\left(x+1\right)-\left(x+1\right)=35\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)=35\)
\(\Rightarrow x^2-1=35\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6;x=-6\)
d)\(x\left(x^2+x+1\right)-x^2\left(x+1\right)=0\)
\(x^3+x^2+x-x^3+x=0\)
\(\Rightarrow x^2+2x=0\)
\(\Rightarrow x\left(x+2\right)=0\)
\(\Rightarrow x=0;x=0-2=-2\)
Vậy \(x=0;x=-2\)
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
Bài 1:
a) \(\left(2x+3\right)\cdot\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3-3=27-3=24\)
--> đpcm
b) Sửa đề: \(\left(x+3\right)^3-\left(x+9\right)\left(x^2+27\right)\)
\(=x^3+9x^2+27x+27-\left(x^3+27x+9x^2+243\right)\)
\(=x^3+9x^2+27x+27-x^3-27x-9x^2-243=27-243=-216\)
--> đpcm
c) \(\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x^2+xy+y^2\right)-2x^3\)
\(=x^3+y^3+x^3-y^3-2x^3=2x^3-2x^3=0\)
--> đpcm
B1: a) \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3-27-8x^3+2\)
\(=-25\)
b) c) Làm theo câu a áp dụng HĐT.
B2:
a) \(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2+3\right)\left(x+2-3\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-5\\x=1\end{matrix}\right..\)
Mấy câu b,c,d bn chịu khó tạo HĐT nhé.
e) \(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=-\dfrac{255}{2}\)
Vậy \(x=-\dfrac{255}{2}\)
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)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
toi ko bt
\(x^2+y^2=0\)
Mà \(x^2\ge0;y^2\ge0\)nên \(x^2+y^2\ge0\)
(Dấu "="\(\Leftrightarrow x=y=0\))
\(x^2+2y^2+2y\left(1-x\right)=-1\)
\(\Leftrightarrow x^2+2y^2+2y-2xy+1=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y+1\right)^2=0\)
Mà \(\left(x-y\right)^2+\left(y+1\right)^2\ge0\)
(Dấu "="\(\Leftrightarrow x=-1;y=-1\)
\(\left(x-1\right)^2+\left(y+2\right)^2=0\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
Dấu "="\(\Leftrightarrow y=-2;x=1\)
\(4x^2+y^2-2\left(2x+y-1\right)=0\)
\(\Leftrightarrow4x^2+y^2-4x-2y+2=0\)
\(\Leftrightarrow\left(2x+1\right)^2+\left(y-1\right)^2=0\)
Mà \(\left(2x+1\right)^2+\left(y-1\right)^2\ge0\)
Dấu "="\(\Leftrightarrow y=1;x=\frac{-1}{2}\)