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\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
Bài 1:
a: \(x^2\left(3x+2\right)=3x^3+2x^2\)
b: \(\left(x-2\right)\left(3x^2-4x+1\right)\)
\(=3x^3-4x^2+x-6x^2+8x-2\)
\(=3x^2-10x^2+9x-2\)
c: \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(x-3\right)\left(x+3\right)\)
\(=27x^3+8-x^2+9=27x^3-x^2+17\)
d: \(=\left(x+y-x-y+z\right)\left(x+y+x+y-z\right)\)
\(=z\left(2x+2y-z\right)\)
\(=2xz+2yz-z^2\)
a) \(x^2-2x=24\)
\(\Rightarrow x^2-2x-24=0\)
\(\Rightarrow x^2-6x+4x-24=0\)
\(\Rightarrow x\left(x-6\right)+4\left(x-6\right)=0\)
\(\Rightarrow\left(x-6\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) \(\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x\right)^2-4^2=0\)
\(\Rightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Rightarrow\left(1-2x\right)\left(9-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1-2x=0\\9-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
c)Sửa đề
\(x^2-4x+4-9x^2+6x-1=0\)
\(\Rightarrow\left(x^2-4x+4\right)-\left(9x^2-6x+1\right)=0\)
\(\Rightarrow\left(x-2\right)^2-\left(3x-1\right)^2=0\)
\(\Rightarrow\left(x-2-3x+1\right)\left(x-2+3x-1\right)=0\)
\(\Rightarrow\left(-2x-1\right)\left(4x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-2x-1=0\\4x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2x=1\\4x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
d) \(2x^2+y^2+2xy-4x+4=0\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(x^2-4x+4\right)=0\)
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2=0\)
Vì \(\left(x+y\right)^2\ge0\) với mọi x và y
\(\left(x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2\ge0\) với mọi x và y
Mà \(\left(x+y\right)^2+\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-x\\x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)
\(•B=4x-9x^2=-\left(9x^2-4x\right)\\ =-\left(9x^2-3x.2.\dfrac{2}{3}+\dfrac{4}{9}\right)+\dfrac{4}{9}\\ =-\left(3x-\dfrac{2}{3}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\\dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{2}{9}\\ vậy\: MAX_B=\dfrac{4}{9}\: tại\: x=\dfrac{2}{9}\\ •C=5-2x-4x^2=-\left(4x^2+2x-5\right)\\ =-\left(4x^2+2.2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{21}{4}\\ =-\left(2x+\dfrac{1}{2}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=-\dfrac{1}{4}\\ vậy\: MAX_C=\dfrac{21}{4}\: tại\: x=\dfrac{-1}{4}\\ •D=7+3x-x^2=-\left(x^2-3x-7\right)\\ =-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{37}{4}\\ =-\left(x-\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{3}{2}\\ vậy\: MAX_D=\dfrac{37}{4}\: tại\: x=\dfrac{3}{2}\)\(•E=1+x-x^2=-\left(x^2-x-1\right)\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\\ dấu\:"="\:xảy\:ra\:khi\:x=\dfrac{1}{2}\\ vậy\:MAX_E=\dfrac{5}{4}\:tại\:x=\dfrac{1}{2}\\ •F=-5x-6x^2\\ -\dfrac{F}{6}=x^2+\dfrac{5}{6}x=x^2+2.x.\dfrac{5}{12}+\dfrac{25}{144}-\dfrac{25}{144}\\ -\dfrac{F}{6}=\left(x+\dfrac{5}{12}\right)^2-\dfrac{25}{144}\\ F=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{25}{24}\le\dfrac{25}{24}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{-5}{12}\\ vậy\: MAX_F=\dfrac{25}{24}\: tại\: x=\dfrac{-5}{12}\)
\(B=4x-9x^2=-9\left(x^2-\dfrac{4}{9}x+\dfrac{4}{81}\right)+\dfrac{4}{9}\)
\(=-9\left(x-\dfrac{2}{9}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\forall x\)
vậy Max B = \(\dfrac{4}{9}\) khi \(x-\dfrac{2}{9}=0\Rightarrow x=\dfrac{2}{9}\)
\(C=5-2x-4x^2=-4\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)+\dfrac{21}{4}\)\(=-4\left(x+\dfrac{1}{4}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\)
Vậy Max C = \(\dfrac{21}{4}\) khi \(x+\dfrac{1}{4}=0\Rightarrow x=-\dfrac{1}{4}\)
\(D=7+3x-x^2\)
\(=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{37}{4}\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\forall x\)
Vậy Max D = \(\dfrac{37}{4}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(E=1+x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{5}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\forall x\)
Vậy Max E = \(\dfrac{5}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
\(F=-5x-6x^2=-6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}\right)+\dfrac{25}{24}\)\(=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{25}{24}\le\dfrac{25}{24}\forall x\)
Vậy Max F = \(\dfrac{25}{24}\) khi \(x+\dfrac{5}{12}=0\Leftrightarrow x=-\dfrac{5}{12}\)
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
a: \(=-\left(x^2+10x-11\right)\)
\(=-\left(x^2+10x+25-36\right)\)
\(=-\left(x+5\right)^2+36< =36\)
Dấu '=' xảy ra khi x=-5
b: \(=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=-\left(x-3\right)^2+4< =4\)
Dấu '=' xảy ra khi x=3
c: \(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
d: \(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9< =9\)
Dấu '=' xảy ra khi x=-1
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
l/ $6x^2(x-1)-9x(x-1)\\=(6x^2-9)(x-1)\\=3(2x^2-3)(x-1)\\=3(\sqrt2 x-\sqrt 3)(\sqrt 2 x+\sqrt 3)(x-1)$
m/ $4x^2(x-2)+9x(2-x)\\=4x^2(x-2)-9x(x-2)\\=(4x^2-9x)(x-2)\\=x(4x-9)(x-2)$
n/ $4x^2y-4xy+y\\=y(4x^2-4x+1)\\=y(2x-1)^2$
o/ $3x(2x-3y)-6(3y-2x)\\=3x(2x-3y)+6(2x-3y)\\=(3x+6)(2x-3y)\\=3(x+2)(2x-3y)$
p/ $4x^2(x-1)+(1-x)\\=4x^2(x-1)-(x-1)\\=(4x^2-1)(x-1)\\=(2x-1)(2x+1)(x-1)$
l)\(6x^2\left(x-1\right)-9x\left(x-1\right)=3x\left(x-1\right)\left(2x-3\right)\)
m) \(4x^2\left(x-2\right)+9x\left(2-x\right)=4x^2\left(x-2\right)-9x\left(x-2\right)=x\left(x-2\right)\left(4x-9\right)\)
n) \(4x^2y-4xy+y=y\left(4x^2-4x+1\right)=y\left(2x-1\right)^2\)
o) \(3x\left(2x-3y\right)-6\left(3y-2x\right)=3x\left(2x-3y\right)+6\left(2x-3y\right)=3\left(2x-3y\right)\left(x+2\right)\)
p) \(4x^2\left(x-1\right)+\left(1-x\right)=4x^2\left(x-1\right)-\left(x-1\right)=\left(4x^2-1\right)\left(x-1\right)=\left(2x-1\right)\left(2x+1\right)\left(x-1\right)\)