Tìm x, biết: a)(x-3)-9.(2x-7)=-16:2 b) -2.(-1+x)+5.(-2+x)=-14+x. Thanks mọi người nhìu nhìu <3
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
19 tháng 6 2017
d) \(x\left(x+1\right)-x-1=0\)
\(\Leftrightarrow x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
16 tháng 9 2017
Pt tương đương:
\(2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(\Leftrightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Leftrightarrow5x=-3\)
\(\Leftrightarrow x=-\frac{3}{5}\)
Vậy pt có nghiệm là :\(x=-\frac{3}{5}\)
14 tháng 5 2016
a) <=> \(2x^2-8x+3x-12+x^2-7x+10=3x^2-5x-12x+20\)
<=> \(2x^2-8x+3x-12+x^2-7x+10-3x^2+5x+12x-20=0\)
<=> \(5x-22=0\)
<=> \(5x=22\)
<=> \(x=\frac{22}{5}\)
b) <=> \(24x^2-9x+16x-6-4x^2-7x-16x-28=10x^2+5x-2x-1\)
<=> \(24x^2-9x+16x-6-4x^2-7x-16x-28-10x^2-5x+2x+1=0\)
<=> \(10x^2-19x-33=0\)
<=> \(10x^2-30x+11x-33=0\)
<=> \(10x\left(x-3\right)+11\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(10x+11\right)=0\)
<=> \(x=3;x=-\frac{11}{10}\)
MC
15 tháng 7 2021
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
MC
15 tháng 7 2021
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
25 tháng 7 2020
1) \(x^2+\frac{8}{9}=\frac{41}{36}\)\(\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)
2) \(\left(x-3\right)^2+-\frac{9}{25}=\frac{2}{5}.\frac{8}{5}\)
\(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
3) \(\frac{3}{11}.\frac{22}{6}-\left(x-1\right)^2=\frac{7}{16}\)
\(\Leftrightarrow\left(x-1\right)^2=\frac{9}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{3}{4}\\x-1=-\frac{3}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=\frac{1}{4}\end{cases}}\)
4) \(1+\left(x+1\right)^3=\frac{37}{64}\)
\(\Leftrightarrow\left(x+1\right)^3=-\frac{27}{64}\)
\(\Rightarrow x+1=-\frac{3}{4}\)
\(\Leftrightarrow x=-\frac{7}{4}\)
5) \(\left(x-\frac{1}{2}\right)^2-\frac{9}{16}=1\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{25}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{5}{4}\\x-\frac{1}{2}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=-\frac{3}{4}\end{cases}}\)
6) Bn ghi rõ đề nha mk ko hiểu
25 tháng 7 2020
6 ) \(\frac{3-x}{5-x}=\left(\frac{-3}{5}\right)^2\)
\(\frac{3-x}{5-x}=\frac{9}{25}\Leftrightarrow\frac{3-x}{5-x}-\frac{9}{25}\Leftrightarrow\frac{75-25x}{125-25x}-\frac{45-9x}{125-5x}=0\)
\(\Rightarrow\frac{75-25-45+9x}{125-25x}=0\Leftrightarrow5+9x=0\Leftrightarrow x=\frac{-5}{9}\)
12 tháng 9 2021
1: Ta có: \(\left(3-x\right)^2+\left(2x+1\right)^2-\left(2-x\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-2\right)=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
2: Ta có: \(\left(1-2x\right)^2-3\left(x-1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow4x^2-4x+1-3x^2+6x-3+\left(x+1\right)^2-2\left(x-1\right)^2=0\)
\(\Leftrightarrow x^2+2x-2+x^2+2x+1-2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow2x^2+4x+1-2x^2+4x-2=0\)
\(\Leftrightarrow x=\dfrac{1}{8}\)
KL
21 tháng 7 2016
a,\(\Leftrightarrow\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)-17=0\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x-17=0\)
\(\Leftrightarrow9x-10=0\)
\(\Leftrightarrow x=\frac{10}{9}\)
3 tháng 7 2018
a. Ta có: \(x^2-10x+26+y^2+2y=0\Leftrightarrow\left(x^2-10x+25\right)+\left(y^2+2y+1\right)=0\\ \)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\Rightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)
b. \(\left(2x+5\right)^2-\left(x-7\right)^2=0\Leftrightarrow\left(2x+5+x-7\right).\left(2x+5-x+7\right)=0\)
\(\Leftrightarrow\left(3x-2\right).\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-12\end{cases}}}\)
c. \(25.\left(x-3\right)^2=49.\left(1-2x\right)^2\Leftrightarrow\left(5x-15\right)^2=\left(7-14x\right)^2\Leftrightarrow\left(5x-15\right)^2-\left(7-14x\right)^2=0\)
\(\Leftrightarrow\left(5x-15-7+14x\right).\left(5x-15+7-14x\right)=0\Leftrightarrow\left(19x-22\right).\left(-9x-8\right)=0\)
\(\Leftrightarrow\left(19x-22\right).\left(9x+8\right)=0\Leftrightarrow\orbr{\begin{cases}19x-22=0\\9x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{22}{19}\\x=-\frac{8}{9}\end{cases}}}\)
d. \(\left(x+2\right)^2=\left(3x-5\right)^2\Leftrightarrow\left(x+2\right)^2-\left(3x-5\right)^2=0\Leftrightarrow\left(x+2+3x-5\right).\left(x+3-3x+5\right)=0\)
\(\Leftrightarrow\left(4x-3\right).\left(8-2x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-3=0\\8-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=4\end{cases}}}\)
e. \(x^2-2x+1=16\Leftrightarrow\left(x-1\right)^2-16=0\Leftrightarrow\left(x-1-4\right).\left(x-1+4\right)=0\)
\(\Leftrightarrow\left(x-5\right).\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
29 tháng 8 2021
Bài 2:
a: Ta có: \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)
\(=\left(x+y\right)^3+2\cdot\left(x+y\right)^2\)
\(=7^3+2\cdot7^2=441\)
a) \(\left(x-3\right)-9\left(2x-7\right)=\dfrac{-16}{2}\)
\(\Rightarrow x-3-18x+63=-8\)
\(\Rightarrow x=4\)
b) \(-2\left(-1+x\right)+5\left(-2+x\right)=-14+x\)
\(\Rightarrow2+-2x-10+5x=-14+x\)
\(\Rightarrow x=-3\)
\(\left(x-3\right)-9.\left(2x-7\right)=-16:2\)
\(\Leftrightarrow\left(x-3\right)-9\left(2x-7\right)=-8\)
\(\Leftrightarrow x-3-18x+63=-8\)
\(\)\(\Leftrightarrow x-3-18x=-8-63\)
\(\Leftrightarrow-17x-3=-71\)
\(\Leftrightarrow-17x=-68\)
\(\Leftrightarrow x=-68:-17=4\)
\(-2\left(-1+x\right)+5\left(-2+x\right)=-14+x\)
\(\Leftrightarrow2-2x-10+5x=-14+x\)
\(\Leftrightarrow-8+3x=-14+x\)
\(\Leftrightarrow3x-x=-14-8\)
\(\Leftrightarrow2x=-6\Rightarrow x=-3\)