K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!
NH
1 tháng 9 2020
\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)
\(\Rightarrow-x=5\)
\(\Rightarrow x=-5\)
\(\text{Vậy x=-5}\)
\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)
\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)
\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)
\(\Rightarrow-16x-8=7\)
\(\Rightarrow-16x=15\)
\(\Rightarrow x=\frac{-15}{16}\)
\(\text{Vậy }x=\frac{-15}{16}\)
\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)
\(\Rightarrow-9+8x-1=8\)
\(\Rightarrow8x=18\)
\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)
\(\text{Vậy }x=\frac{9}{4}\)
\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)
14 tháng 10 2021
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
NM
4
2 tháng 9 2021
a) \(3\left(x-2\right)+2\left(x-3\right)=5\)
\(\Rightarrow3x-6+2x-6=5\)
\(\Rightarrow5x=17\Rightarrow x=\dfrac{17}{5}\)
b) \(\left(2x-8\right)^2-16=0\)
\(\Rightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)
\(\Rightarrow\left(2x-12\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=12\\2x=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
c) \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)
\(\Rightarrow4x^2-4x+1-4x^2+12x-x+3=3\)
\(\Rightarrow7x=-1\Rightarrow x=-\dfrac{1}{7}\)
2 tháng 9 2021
a: Ta có: \(3\left(x-2\right)+2\left(x-3\right)=5\)
\(\Leftrightarrow3x-6+2x-6=5\)
\(\Leftrightarrow5x=17\)
hay \(x=\dfrac{17}{5}\)
b: Ta có: \(\left(2x-8\right)^2-16=0\)
\(\Leftrightarrow\left(2x-4\right)\left(2x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
CM
5
8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
25 tháng 8 2019
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=0+4\)
\(\left(x-3\right)^2=4\)
\(\left(x-3\right)^2=\pm4\)
\(\left(x-3\right)^2=\pm2^2\)
\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(x=1\)
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
\(16x^2-9-16x^2+40x-25=16\)
\(-34+40x=16\)
\(40x=16+34\)
\(40x=50\)
\(x=\frac{50}{40}=\frac{5}{4}\)
d) \(x^3-9x^2+27x-27=-8\)
\(x^3-9x^2+27x-27+8=0\)
\(x^3-9x^2+27x-19=0\)
\(\left(x^2-8x+19\right)\left(x-1\right)=0\)
Vì \(\left(x^2-8x+19\right)>0\) nên:
\(x-1=0\)
\(x=1\)
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)
\(3x+1=2\)
\(3x=2-1\)
\(3x=1\)
\(x=\frac{1}{3}\)
CD
1
14 tháng 9 2021
1) \(\dfrac{3x}{4x-8}\)
\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)
2) \(\dfrac{2x}{x^2-9}\)
\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)
(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)
\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)
5) \(\dfrac{x-2}{x^2+3}\)
Do \(x^2+3>0\forall x\in R\)
Vậy biểu thức trên xác định với mọi x
6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)
\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)
\(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow x-3=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)\left[1-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
___________
\(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)