Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ANH HAY CHỊ ƠI LÀM GIÚP EM BAI LỚP 7 ĐI O DUOI DAY A
a) \(\left(x-3\right)^2-4=0\)
\(\Rightarrow\left(x-3\right)^2=4\)
\(\Rightarrow\left(x-3\right)^2=2^2=\left(-2\right)^2\)
\(\Rightarrow x-3=2\)hoặc \(\left(x-3\right)=-2\)
\(\Rightarrow\hept{\begin{cases}x-3=2\\x-3=-2\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=-1\end{cases}}}\)
Vậy \(x\in\left\{5;-1\right\}\)
b) \(x^2-2x=24\)
\(\Rightarrow x.\left(x+2\right)=24\)
\(\Rightarrow x.\left(x+2\right)=4.6\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
a. \(9\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow9x+18-3x-6=0\)
\(\Leftrightarrow6x+12=0\)
\(\Leftrightarrow x=-2\)
e. \(\left(2x-1\right)^2-45=0\)
\(\Leftrightarrow4x^2-2x+1-45=0\)
\(\Leftrightarrow4x^2-2x-44=0\)
Đến đó tự giải tiếp nha!
c. \(2\left(2x-5\right)-3x=0\)
\(\Leftrightarrow4x-10-3x=0\)
\(\Leftrightarrow x-10=0\)
\(\Leftrightarrow x=10\)
g. \(2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
a) \(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left(x+2\right)^2=3^2\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=3-2=1\)
a) ( x + 2 )2 = 9
=> ( x + 2 ) 2 = 9
=> ( x + 2 )2 = 32
=> x + 2 = + 3
=> \(\orbr{\begin{cases}x+2=-3\\x+2=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
Vậy x = -1; 5
b) ( x + 2 )2 - x2 + 4 = 0
=> ( x + 2 )2 - ( x2 - 4 ) = 0
=> ( x + 2 )2 - ( x + 2 ) ( x - 2 ) = 0
=> ( x + 2 ) ( x + 2 - x + 2 ) = 0
=> ( x + 2 ) . 4 = 0
=> x + 2 = 0
=> x = - 2
Vậy x = - 2
c) 5 ( 2x - 3 )2 - 5 ( x + 1 )2 - 15( x + 4 ) ( x - 4 ) = - 10
=> 5 ( 4x2 - 12x + 9 ) - 5 ( x2 + 2x + 1 ) - 15 ( x2 - 42 ) = - 10
=> 20x2 - 60x + 45 - 5x2 - 10x - 5 - 15x2 + 240 = -10
=> - 70x + 280 = - 10
=> - 70x = - 290
=> x = \(\frac{29}{7}\)
Vậy x = \(\frac{29}{7}\)
d) x ( x + 5 ) ( x - 5 ) - ( x + 2 ) ( x2 - 2x + 4 ) = 3
=> x ( x2 - 25 ) - ( x3 - 8 ) = 3
=> x3 - 25x - x3 + 8 = 3
=> - 25x + 8 = 3
=> - 25x = -5
=> x = \(\frac{1}{5}\)
Vậy x = \(\frac{1}{5}\)
a ) \(\left(5x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{5}\\x=7\end{matrix}\right.\)
b ) \(\left(x^2-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-3\end{matrix}\right.\)
c )\(x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
d ) \(x^2+x-12=0\)
\(\Leftrightarrow x^2-4x+3x-12\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e ) \(15\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=-21\end{matrix}\right.\)
g ) \(\left(x^2+1\right)\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-2\end{matrix}\right.\)
i ) \(x^4+2x^3-2x^2+2x-3=0\)
\(\Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\)
\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^3-x^2+x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=1\\x=-3\end{matrix}\right.\)
h) \(x^2+5x+6=0\)
\(\Leftrightarrow x^2+3x+2x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Bài giải:
a) x3 – 1414x = 0 => x(x2 –(12)2(12)2) = 0
=>x(x - 1212)(x + 1212) = 0
Hoặc x = 0
Hoặc x - 1212 = 0 => x = 1212
Hoặc x + 1212 = 0 => x = -1212
Vậy x = 0; x = -1212; x = 1212.
b) (2x – 1)2 – (x + 3)2 = 0
[(2x - 1) - (x + 3)][(2x - 1) + (x + 3)] = 0
(2x - 1 - x - 3)(2x - 1 + x + 3) = 0
(x - 4)(3x + 2) = 0
Hoặc x - 4 = 0 => x = 4
Hoặc 3x + 2 = 0 => 3x = 2 => x = -2323
Vậy x = 4; x = -2323.
c) x2(x – 3) + 12 – 4x = 0
x2(x – 3) - 4(x -3)= 0
(x - 3)(x2- 22) = 0
(x - 3)(x - 2)(x + 2) = 0
Hoặc x - 3 = 0 => x = 3
Hoặc x - 2 =0 => x = 2
Hoặc x + 2 = 0 => x = -2
Vậy x = 3; x = 2; x = -2.
a ) \(x^3-\dfrac{1}{4}x=0\)
\(\Leftrightarrow\) \(x\left(x^2-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
Hoặc x = 0
Hoặc \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
Hoặc \(x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{2}\)
b) \((2x - 1 )^2 - (x + 3)^2 = 0\)
\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x-3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)
Hoặc \(x-4=0\Rightarrow x=4\)
Hoặc \(3x+2=0\Rightarrow3x=-2\Rightarrow x=-\dfrac{2}{3}\)
c) \(x^2 (x-3) + 12 - 4x = 0\)
\(\Leftrightarrow x^2\left(x-3\right)-\left(4x-12\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-2^2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)=0\)
Hoặc \((x - 3) = 0\) \(\Rightarrow\) x = 3
Hoặc \(x - 2 = 0\) \(\Rightarrow\) x = 2
Hoặc \(x + 2 = 0 \) \(\Rightarrow\) x = \(- 2\)
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) 4x.(x+1) = 8.(x+1)
=> 4x.(x+1) - 8.(x+1) = 0
4.(x+1).(x-2) = 0
=> x + 1 = 0 => x = -1
x-2 = 0 => x = 2
KL:...
b) x.(x-1) - 2.(1-x) = 0
x.(x-1) + 2.(x-1) = 0
(x-1).(x+2) = 0
...
c) 2x.(x-2) - (2-x)2 = 0
2x.(x-2) - (x-2)2 = 0
(x-2).( 2x - x + 2) = 0
(x-2).(x+2) = 0
...
d) (x-3)3 + 3 -x = 0
(x-3)3 - (x-3) = 0
(x-3).[(x-3)2 - 1] = 0
(x-3).(x-3+1).(x-3-1) = 0
(x-3).(x-2).(x-4)=0
=> ...
\(x^3-2x^2-x+2=0\)
\(\Rightarrow x^2\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
Tìm được \(x\in\left\{2;1;-1\right\}\)
\(\left(x^2+x\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)=0\)(1)
Mà \(x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\) (2)
Từ (1) và (2) \(\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)