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27 tháng 9 2020
a) \(3x^3-12x=0\)
=> \(3x\left(x^2-4\right)=0\)
=> \(\orbr{\begin{cases}3x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
=> \(x^2\left(x-3\right)+\left(-4x+12\right)=0\)
=> \(x^2\left(x-3\right)-4x+12=0\)
=> \(x^2\left(x-3\right)-4\left(x-3\right)=0\)
=> \(\left(x-3\right)\left(x^2-4\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
=> \(\left[3x-1-\left(2x-3\right)\right]\left(3x-1+2x-3\right)=0\)
=> \(\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\)
=> \(\left(x+2\right)\left(5x-4\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{4}{5}\end{cases}}\)
d) \(x^2-4x-21=0\)
=> \(x^2+3x-7x-21=0\)
=> \(x\left(x+3\right)-7\left(x+3\right)=0\)
=> (x + 3)(x - 7) = 0 => x = -3 hoặc x = 7
e) 3x2 - 7x - 10 = 0
=> 3x2 + 3x - 10x - 10 = 0
=> 3x(x + 1) - 10(x + 1) = 0
=> (x + 1)(3x - 10) = 0
=> x = -1 hoặc x = 10/3
F
27 tháng 9 2020
a) \(3x^3-12x=0\)
\(\Leftrightarrow3x\left(x^2-4\right)=0\)
\(\Leftrightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2;0;2\right\}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\in\left\{-2;2;3\right\}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5x-4\right)=0\)
\(\Leftrightarrow x\in\left\{-2;\frac{4}{5}\right\}\)
XO
27 tháng 9 2020
Ta có : 3x3 - 12x = 0
=> 3x(x2 - 4) = 0
=> x(x - 2)(x + 2) = 0
=> \(x\in\left\{0;2;-2\right\}\)
b) x2(x - 3) + 12 - 4x = 0
=> x2(x - 3) - 4(x - 3) = 0
=> (x2 - 4)(x - 3) = 0
=> \(\orbr{\begin{cases}x^2-4=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm2\\x=3\end{cases}}\)
Vậy \(x\in\left\{-2;2;3\right\}\)
c) (3x - 1)2 - (2x - 3)2 = 0
=> (3x - 1 - 2x + 3)(3x - 1 + 2x - 3) = 0
=> (x + 2)(5x - 4) = 0
=> \(\orbr{\begin{cases}x+2=0\\5x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=0,8\end{cases}}\)
Vậy \(x\in\left\{-2;0,8\right\}\)
d) x2 - 4x - 21 = 0
=> x2 - 7x + 3x - 21 = 0
=> x(x - 7) + 3(x - 7) = 0
=> (x + 3)(x - 7) = 0
=> \(\orbr{\begin{cases}x+3=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=7\end{cases}}\)
Vậy \(x\in\left\{-3;7\right\}\)
e) 3x2 - 7x - 10 = 0
=> 3x2 + 3x - 10x - 10 = 0
=> 3x(x + 1) - 10(x + 1) = 0
=> (3x - 10)(x + 1) = 0
=> \(\orbr{\begin{cases}3x-10=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=-1\end{cases}}\)
Vậy \(x\in\left\{\frac{10}{3};-1\right\}\)
VQ
3
4 tháng 8 2019
\(a,-2x\left(2-3x\right)+3\left(-5+7x-6x^2\right)=-4\)
\(\Rightarrow-4x+6x^2-15+21x-18x^2=-4\)
\(\Rightarrow-12x^2+17x-11=0\)
\(\Rightarrow12x^2-17x+11=0\)
\(\Rightarrow9x^2-2.3.\frac{17}{6}x+\left(\frac{17}{6}\right)^2-\left(\frac{17}{6}\right)^2+11=0\)
\(\Rightarrow\left(3x-\frac{17}{6}\right)^2+\frac{107}{36}=0VN\)
Không có gt x thỏa mãn
\(b,-3x\left(-1+3x-4x^2\right)+6x^2\left(-2x+3\right)=0\)
\(\Rightarrow3x-9x^2+12x^3-12x^3+18x^2=0\)
\(\Rightarrow9x^2+3x=0\)
\(\Rightarrow3x\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\3x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\frac{1}{3}\end{cases}}}\)
LR
1
27 tháng 12 2015
thêm câu b, x2-7x+10
=x2-2x-5x+10
=(x2-2x)-(5x+10)
= x(x-2)-5(x+2)
=(x-2)(x-5)
HT
2
D
5 tháng 8 2019
a, 4x2 + 4x + 1 = 25
⇔ ( 2x + 1 ) \(^2\) - 25 = 0
⇔( 2x + 1 - 5 ) ( 2x + 1 + 5 ) =0
⇔ ( 2x - 4 ) ( 2x + 6 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
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:
\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
TQ
4 tháng 3 2018
\(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow\left(2x^3+7x^2+7x\right)+2=0\)
\(\Leftrightarrow x\left(2x^2+7x+7+2\right)=0\)
\(\Leftrightarrow x\left(2x^2+7x+9\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x+3x+9\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2+6x\right)+\left(3x+9\right)\right]=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{3}{2}\end{matrix}\right.\)
chúc bạn học tốt!
TT
1
LC
6 tháng 5 2017
a,A= x(x3-5x2+7x-3)
=x(x3-3x2-2x2+6x+x-3)
=x(x-3)(x2-2x+1)
=x(x-3)(x-1)2
vi (x-1)2>=0
=>Để A <0 thì x(x-3)<0
TH1:x>0 va x-3<0
x>0 va x<3
=> 0<x<3
TH2 :x<0 va x-3>0
x<0 và x>3( loại vỉ 2 dk trái ngược nhau )
Vay 0<x<3 thi thoa man....... .........
Phần b tương tự