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27 tháng 7 2018
Ukm
It's very hard
l can't do it
Sorry!
27 tháng 7 2018
a) \(x^4-x^3-7x^2+x+6=0\)
\(\Leftrightarrow x^4+2x^3-3x^3-6x^2-x^2-2x+3x+6=0\)
\(\Leftrightarrow x^3\left(x+2\right)-3x^2\left(x+2\right)-x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-3x^2-x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-3\right)=0\). Làm nốt
b) \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)
\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-3\right)^2+\left|y+3\right|=0\)
Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)
\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)
c) \(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)^2-2.\left(2x^2+x\right).2+4-1=0\)
\(\Leftrightarrow\left(2x^2+x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}2x^2+x-2=1\\2x^2+x-2=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)
Từ đó tính đc x
d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(x^2+5x+5=a\), khi đó pt có dạng:
\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
A
1 tháng 9 2020
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
18 tháng 7 2019
2 .tìm x
a , x ( x + 2 ) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b, x ( x-5 )= 5 -x
<=> x ( x-5 ) + x - 5 = 0
<=> x (x-5) + ( x-5)= 0
<=> (x-5)(x+1 )=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
c) ( x + 1 ) ( 6x2 + 2x ) + ( x - 1 ) ( 6x2 + 2x ) = 0
\(\Leftrightarrow\) ( 6x2 + 2x ) \([\)(x+1)(x-1)\(]\)=0
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x\left(3x+1\right)=0\\x^{2^{ }}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3x+1=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\frac{-1}{3}\\x=1\end{matrix}\right.\)
18 tháng 7 2019
1 ,a) 2a ( x - y ) - ( y - x ) = 2ax - 2ay - y + x
= x ( 2a + 1 ) - y ( 2a + 1 )
= ( 2a + 1 ) ( x - y )
b) a2 ( x - y ) - ( y - x ) = a2x - a2y - y + x
= x ( a2+ 1 ) - y ( a2 +1 )
= ( a2+1 ) - (x-y )
c) x ( x - y ) + y ( y - x ) - 3 ( x - y ) = x 2 - xy -+ y 2 - xy - 3x + 3y
= x2 - 2xy + y2 -3x + 3y
= (x-y)2 -3 ( x - y )
= ( x-y ) ( x-y+3)
13 tháng 7 2015
1a/ x3+x2+x+1=0
x2(x+1).(x+1)=0
=> x2(x+1)=0 x =1
hoặc =>[
x+1=0 x=-1
b/(x+2)2=x+2
x2+2.x.2+22 =x+2
x+x+4x+4=x+2
6x+4=x+2
....
c/(x+1)(6x2+2x)+(x-1)(6x2+2x)=0
x2-12 + (6x2+2x)2=0
=> x2-1 = 0 x=1
hoặc => [
(6x2+2x)2=0 x= 0
17 tháng 9 2018
Bài 1 :
\(a)\)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+3\right)\left(x-3\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x\left(x^2-3^2\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x^3+9x=15\)
\(\Leftrightarrow\)\(9x=16\)
\(\Leftrightarrow\)\(x=\frac{16}{9}\)
Vậy \(x=\frac{16}{9}\)
Chúc bạn học tốt ~
M
30 tháng 10 2019
Câu 1 : Tìm x :
1. \(A=x^2+4x-2\)
\(A=x^2+2.x.2+2^2-2^2-2\)
\(A=\left(x^2+4x+2^2\right)-4-2\)
\(A=\left(x+2\right)^2-6\)
\(\left(x+2\right)^2-6\ge-6\)
MIn A= -6 khi \(\left(x+2\right)^2=0\)
=> \(x+2=0hayx=-2\)
Vậy x=2
những câu tiếp theo làm tg tự như thế nhé
30 tháng 10 2019
Câu 1:
a) Ta có: \(A=x^2+4x-2\)
\(=x^2+4x+4-6\)
\(=\left(x+2\right)^2-6\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2-6\ge-6\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: x=-2
b) Ta có: \(B=2x^2-4x+3\)
\(=2\left(x^2-2x+\frac{3}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot1+1+\frac{1}{2}\right)\)
\(=2\left[\left(x^2-2x\cdot1+1\right)+\frac{1}{2}\right]\)
\(=2\left[\left(x-1\right)^2+\frac{1}{2}\right]\)
\(=2\left(x-1\right)^2+1\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy: x=1
c) Ta có: \(C=x^2+y^2-4x+2y+5\)
\(=x^2-4x+4+y^2+2y+1\)
\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\left(y+1\right)^2\ge0\forall y\)
Do đó: \(\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy: x=2 và y=-1
Câu 2:
a) Ta có: \(A=-x^2+6x+5\)
\(=-\left(x^2-6x-5\right)\)
\(=-\left(x^2-6x+9-14\right)\)
\(=-\left[\left(x^2-6x+9\right)-14\right]\)
\(=-\left[\left(x-3\right)^2-14\right]\)
\(=-\left(x-3\right)^2+14\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2+14\le14\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(A=-x^2+6x+5\) là 14 khi x=3
b) Ta có: \(B=-4x^2-9y^2-4x+6y+3\)
\(=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(4x^2+4x+1\right)+\left(9y^2-6y+1\right)-5\right]\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2-5\right]\)
\(=-\left(2x+1\right)^2-\left(3y-1\right)^2+5\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow-\left(2x+1\right)^2\le0\forall x\)(1)
Ta có: \(\left(3y-1\right)^2\ge0\forall y\)
\(\Rightarrow-\left(3y-1\right)^2\le0\forall y\)(2)
Từ (1) và (2) suy ra
\(-\left(2x+1\right)^2-\left(3y-1\right)^2\le0\forall x,y\)
\(\Rightarrow-\left(2x+1\right)^2-\left(3y-1\right)^2+5\le5\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}-\left(2x+1\right)^2=0\\-\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\3y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\3y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)
Vậy: GTLN của đa thức \(B=-4x^2-9y^2-4x+6y+3\) là 5 khi và chỉ khi \(x=\frac{-1}{2}\) và \(y=\frac{1}{3}\)
Câu 3:
a) Ta có: \(x^2+y^2-2x+4y+5=0\)
\(\Rightarrow x^2-2x+1+y^2+4y+4=0\)
\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy: x=1 và y=-2
b) Ta có: \(5x^2+9y^2-12xy-6x+9=0\)
\(\Rightarrow x^2+4x^2+9y^2-12xy-6x+9=0\)
\(\Rightarrow\left(4x^2+12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Rightarrow\left(2x+3y\right)^2+\left(x-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot3+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Vậy: x=3 và y=-2
I don't now
or no I don't
..................
sorry
a) \(x^4-x^3-7x^2+x+6=0\)
\(\Leftrightarrow\)\(x^4-x^3-7x^2+7x-6x+6=0\)
\(\Leftrightarrow\)\(x^3\left(x-1\right)-7x\left(x-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^3-7x-6\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=0\)
đến đây lm tiếp