Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tìm x
a) 9(3x-2)=x(2-3x)
b) 25x2-2=0
c) x2-25=6x-9
d) (x+2)2-(x-2)(x+2)=0
e) x3-8=(x-2)3
f) x3+5x2-4x-20=0
a) 9(3x - 2) = x(2 - 3x)
\(\Leftrightarrow\)-9(2 - 3x) = x(2 - 3x)
\(\Leftrightarrow\)-9(2 - 3x) - x(2 - 3x) = 0
\(\Leftrightarrow\)(2 - 3x)(- 9 - x) = 0
\(\Leftrightarrow\)2 - 3x = 0 hay - 9 - x = 0
\(\Leftrightarrow\) 3x = 2 \(\Leftrightarrow\) x = - 9
\(\Leftrightarrow\) x = 2/3
b) 25x2 - 2 = 0
\(\Leftrightarrow\)(5x)2 - (\(\sqrt{2}\))2 = 0
\(\Leftrightarrow\)(5x - \(\sqrt{2}\))(5x + \(\sqrt{2}\)) = 0
\(\Leftrightarrow\)5x - \(\sqrt{2}\)= 0 hay 5x + \(\sqrt{2}\)= 0
\(\Leftrightarrow\)5x = \(\sqrt{2}\) \(\Leftrightarrow\)5x = -\(\sqrt{2}\)
\(\Leftrightarrow\) x = \(\sqrt{2}\)/5 \(\Leftrightarrow\) x = -\(\sqrt{2}\)/5
c) x2 - 25 = 6x - 9
\(\Leftrightarrow\)(x2 - 6x + 9) - 25 = 0
\(\Leftrightarrow\)(x - 3)2 - 52 = 0
\(\Leftrightarrow\)(x - 3 - 5)(x - 3 + 5) = 0
\(\Leftrightarrow\)(x - 7)(x + 2) = 0
\(\Leftrightarrow\)x - 7 = 0 hay x + 2 = 0
\(\Leftrightarrow\)x = 7 \(\Leftrightarrow\)x = -2
d) (x + 2)2 - (x - 2)(x + 2) = 0
\(\Leftrightarrow\)(x + 2)(x + 2) - (x - 2)(x + 2) = 0
\(\Leftrightarrow\)(x + 2)(x + 2 - x + 2) = 0
\(\Leftrightarrow\)(x + 2)4 = 0 (hay 4(x + 2) = 0)
\(\Leftrightarrow\)x + 2 = 0 (vì 4 \(\ne\)0)
\(\Leftrightarrow\)x = -2
e) x3 - 8 = (x - 2)3
\(\Leftrightarrow\)x3 - 23 = (x - 2)3
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4) = (x - 2)3
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4) - (x - 2)3 = 0
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4) - (x - 2)(x - 2)2 = 0
\(\Leftrightarrow\)(x - 2)[x2 + 2x + 4 - (x - 2)2] = 0
\(\Leftrightarrow\)(x - 2)[x2 + 2x + 4 - (x2 - 4x + 4)] = 0
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4 - x2 + 4x - 4) = 0
\(\Leftrightarrow\)(x - 2)6x = 0 (hay 6x(x - 2) = 0)
\(\Leftrightarrow\)x - 2 = 0 hay x = 0 (vì 6\(\ne\)0)
\(\Leftrightarrow\)x = 2
f) x3 + 5x2 - 4x - 20 = 0
\(\Leftrightarrow\)x2(x + 5) - 4(x + 5) = 0
\(\Leftrightarrow\)(x + 5)(x2 - 4) = 0
\(\Leftrightarrow\)(x + 5)(x - 2)(x + 2) = 0
\(\Leftrightarrow\)x + 5 = 0 hay x - 2 = 0 hay x + 2 = 0
\(\Leftrightarrow\)x = -5 \(\Leftrightarrow\)x = 2 \(\Leftrightarrow\)x = -2
b) \(x^3-5x^2+4x-20=0\)
\(=\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(=x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(=\left(x^2+4\right)\left(x-5\right)=0\)
\(x^2\ge0\)
\(\Rightarrow x^2+4\ge4>0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
a. 5x.(12x+7)-3x.(20x-5)=-150
x=-3
b. ( 2x-1).(3-x)+(x+4).(2x-5)=20
x=43/10
c. 9x2-1+(3x-1)2=0
x=1/3
d. 3x.(x-2)-(3x+2).(x-1)=7
x=-5/2
e. (2x-1)2-(2x+5).(2x-5)=20
x=3/2
f. 4x2-5=4
x=3/2
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
<=>\(\left(x^3-4x^2\right)+\left(x^2-4x\right)+\left(5x-20\right)=0\)
<=>\(x^2\left(x-4\right)+x\left(x-4\right)+5\left(x-4\right)=0\)
<=>\(\left(x^2+x+5\right)\left(x-4\right)=0\)
Vì \(x^2+x+5>0\)=>x-4=0
<=>x=4