(3x - 4)2 - 36=0. Tìm x
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
KN
2 tháng 8 2019
a) \(x^2-36=0\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)
KN
2 tháng 8 2019
b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)
\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)
\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)
MA
1
B
30 tháng 10 2020
a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)
\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)
\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)
b) \(3x\left(5-x\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)
c) \(x\left(x-3\right)+4x-12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
d) \(x^2-36=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
e) \(x^2+3x+1=2\)
\(\Leftrightarrow x^2+3x+1-2=0\)
\(\Leftrightarrow x^2+3x-1=0\)
\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)
Còn lại ........... Tự lm nất nha
Bài 1.
4 tháng 1 2022
a: =>3x+10-2x=0
hay x=-10
c: \(\Leftrightarrow3x^2-3x^2+6x=36\)
=>6x=36
hay x=6
12 tháng 12 2021
\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)
Vậy: ...
\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy: ...
\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
Vậy: ...
\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)
Vậy: ...
\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
Vậy: ...
\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)
Vậy: ...
12 tháng 12 2021
câu c sao tính ra vậy đc vậy k hiểu giải thích hộ e đi 36 đâu mất òi
Y
7 tháng 7 2023
\(5,4x^2-36=0\\ \Leftrightarrow\left(2x\right)^2-6^2=0\\ \Leftrightarrow\left(2x-6\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{3;-3\right\}\)
\(7,\left(3x+1\right)^2-16=0\\ \Leftrightarrow\left(3x+1\right)^2-4^2=0\\ \Leftrightarrow\left(3x+1-4\right)\left(3x+1+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\3x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(S=\left\{1;-\dfrac{5}{3}\right\}\)
\(8,\left(2x-3\right)^2-49=0\\ \Leftrightarrow\left(2x-3\right)^2-7^2=0\\ \Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-10=0\\2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-2;5\right\}\)
2 tháng 11 2021
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
\(\left(3x-4\right)^2-36=0\)
\(\left(3x-4\right)^2-6^2=0\)
\(\left(3x-4-6\right)\left(3x-4+6\right)=0\)
\(\left(3x-10\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-10=0\\3x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{10}{3}\\x=-\frac{2}{3}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=\frac{10}{3}\\x=-\frac{2}{3}\end{cases}}\)