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\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
a. \(x^2+3x+5\)
\(=x^2+2.x^2.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
=> đpcm
1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
Bài 1:Tìm x,y biết:
a)\(x^2-6x+y^2+10y+34\)
=>\(\left(x^2-2.x.3+3^2\right)+\left(y^2+2.y.5+5^2\right)=0\)
=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)
=>\(\left\{{}\begin{matrix}x-3=0\\y+5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@
B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
a) Ta có: \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\frac{\left(2x+1\right)^2\cdot3}{15}-\frac{5\left(x-1\right)^2}{15}-\frac{7x^2-14x-5}{15}=0\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)-7x^2+14x+5=0\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x+3=0\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=\frac{-3}{36}\)
Vậy: \(x=\frac{-3}{36}\)
b) Ta có: \(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{201-x}{99}+\frac{203-x}{97}-\frac{205-x}{95}-3=0\)
\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{201-x+99}{99}+\frac{203-x+97}{97}+\frac{205-x+95}{95}=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\)
nên 300-x=0
\(\Leftrightarrow x=300\)
Vậy: x=300
c) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+1\ge1\ne0\forall x\)(2)
Từ (1) và (2) suy ra x+1=0
hay x=-1
Vậy: x=-1
d) Ta có: \(\left(x-1\right)x\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt \(x^2+x-1=t\)
\(\Leftrightarrow\left(t+1\right)\left(t-1\right)=24\)
\(\Leftrightarrow t^2-1-24=0\)
\(\Leftrightarrow t^2-25=0\)
\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-1-5\right)\left(x^2+x-1+5\right)=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{15}{4}\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\right]\)(3)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\ne0\forall x\)(4)
Từ (3) và (4) suy ra
\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
e) Ta có: \(\left(5x-3\right)-\left(4x-7\right)=0\)
\(\Leftrightarrow5x-3-4x+7=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy: x=-4
f) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{1}{3}\right\}\)
g) Ta có: \(x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-8\right\}\)
h) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;2\right\}\)
i) Ta có: \(x^2+x-2=0\)
\(\Leftrightarrow x^2-x+2x-2=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-2\right\}\)
k) Ta có: \(3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;\frac{-1}{3}\right\}\)
l) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-2x-10x+5=0\)
\(\Leftrightarrow2x\left(2x-1\right)-5\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
b) \(7x\left(x-2\right)-\left(x-2\right)=0\)
<=> \(\left(7x-1\right)\left(x-2\right)=0\)
=> x=1/7 hoặc x=2
c) <=> (2x-1)3 =0
=> x=1/2
d)<=> \(\left(2x-3\right)\left(2x+3\right)-x\left(2x-3\right)=0\)
<=> \(\left(2x-3\right)\left(x+3\right)=0\)
=> x=3/2 hoặc x=-3
e) <=>\(x^2\left(x+5\right)+9\left(x+5\right)=0\)
<=> \(\left(x+5\right)\left(x^2+9\right)=0\)
=> x=-5
f) \(x^3-6x^2-x+30=0\)
<=>\(x^3+2x^2-8x^2-16x+15x+30=0\)
<=>\(x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)
<=>\(\left(x+2\right)\left(x^2-5x-3x+15\right)=0\)
<=> \(\left(x+2\right)\left(x-5\right)\left(x-3\right)=0\)
=> x=-2 hoặc x=5 hoặc x=3
a) Ta có : x3 - x = 0
=> x(x2 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
Vậy \(x\in\left\{0;1;-1\right\}\)
b) x2 + 4x = 0
=> x(x + 4) = 0
=> \(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
Vậy \(x\in\left\{0;-4\right\}\)
c) 9x2 - 1 = 0
=> 9x2 = 1
=> x2 = \(\frac{1}{9}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{3};-\frac{1}{3}\right\}\)
d) 5x2 - 10x + 5 = 0
=> 5x2 - 5x - 5x + 5 = 0
=> 5x(x - 1) - 5(x - 1) = 0
=> 5(x - 1)2 = 0
=> (x - 1)2 = 0
=> x - 1 = 0
=> x = 1
e) x2 + 6x + 5 = 0
=> x2 + 6x + 9 - 4 = 0
=> (x + 3)2 = 4
=> \(\orbr{\begin{cases}x+3=2\\x+3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)
Vậy \(x\in\left\{-1;-5\right\}\)
a, \(x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b, \(5x^2-10x+5=0\)
\(\Leftrightarrow5x\left(x-1\right)^2=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)