Tìm x, biết:
a) 8 x 3 - 50x = 0;
b) 2(x + 3)- x 2 - 3x = 0;
c) 6 x 2 - 15x - (2x - 5)(2x + 5) = 0.
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
CM
Cao Minh Tâm
28 tháng 12 2017
Đúng(0)
Những câu hỏi liên quan
10 tháng 10 2021
a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
29 tháng 7 2021
a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
29 tháng 7 2021
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
13 tháng 12 2023
a: \(3\left(x-3\right)-6x=0\)
=>\(3x-9-6x=0\)
=>-3x-9=0
=>3x+9=0
=>3x=-9
=>\(x=-\dfrac{9}{3}=-3\)
b: Đề thiếu vế phải rồi bạn
c: \(2\left(x-3\right)+3x=9\)
=>2x-6+3x=9
=>5x-6=9
=>5x=6+9=15
=>x=15/5=3
d: \(x\left(x-11\right)+2\left(x-11\right)=0\)
=>\(\left(x-11\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)
e: \(x\left(x+2\right)+8=x^2\)
=>\(x^2+2x+8=x^2\)
=>2x+8=0
=>2x=-8
=>x=-8/2=-4
f: \(8\left(x+1\right)+2x=-2\)
=>\(8x+8+2x=-2\)
=>10x=-2-8=-10
=>\(x=-\dfrac{10}{10}=-1\)
g: 12-3(x+2)=0
=>3(x+2)=12
=>x+2=12/3=4
=>x=4-2=2
LH
0
6 tháng 12 2023
\(a,1-3\left|2x-3\right|=-\dfrac{1}{2}\\ 3\left|2x-3\right|=1+\dfrac{1}{2}\\ 3\left|2x-3\right|=\dfrac{3}{2}\\ \left|2x-3\right|=\dfrac{3}{2}:3\\ \left|2x-3\right|=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy `x in {15/4;-3/4}`
\(b,\left(\left|x\right|-0,2\right)\left(x^3-8\right)=0\\ \left(\left|x\right|-0,2\right)\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|-0,2=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|=0,2\\x=2\\\left(x+1\right)^2+3=0\left(lọai\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\\x=2\end{matrix}\right.\)
Vậy `x in {+-0,2;2}`
HL
2
7 tháng 10 2021
a) \(3\sqrt{x-3}=12\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{x-3}=4\)
\(\Leftrightarrow x-3=16\Leftrightarrow x=19\left(tm\right)\)
b) \(\sqrt{16\left(1-2x\right)}-8=0\left(đk:x\le\dfrac{1}{2}\right)\)
\(\Leftrightarrow4\sqrt{1-2x}=8\Leftrightarrow\sqrt{1-2x}=2\)
\(\Leftrightarrow1-2x=4\Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)
c) \(\sqrt{4\left(9-6x+x^2\right)}-12=0\)
\(\Leftrightarrow2\sqrt{\left(x-3\right)^2}=12\)
\(\Leftrightarrow\left|x-3\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=6\\x-3=-6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
7 tháng 10 2021
a: ta có: \(3\sqrt{x-3}=12\)
\(\Leftrightarrow x-3=16\)
hay x=19
b: Ta có: \(\sqrt{16\left(1-2x\right)}-8=0\)
\(\Leftrightarrow1-2x=4\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
AL
4
25 tháng 10 2021
\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)
25 tháng 10 2021
a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)
c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)
