\(\left(2x-1\right)^2-49-4x^2=0\)

">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

15 tháng 1 2022

\(\left(2x-1\right)^2-49-4x^2=0.\Leftrightarrow4x^2-4x+1-49-4x^2=0.\Leftrightarrow-4x=48.\Leftrightarrow x=-12.\)

\(\Leftrightarrow4x^2-4x+1-49-4x^2=0\)

=>4x+48=0

hay x=-12

3 tháng 4 2020

a) 4         b) 0,6 hoặc 1,75     e) -3,5 hoặc -3

29 tháng 6 2017

a) \(\left(3x-5\right)\left(9x^2+15x+25\right)\)

\(=\left(3x\right)^3-5^3\)

\(=27x^3-125\)

b) \(\left(2x+7\right)\left(x^2-14x+49\right)-2x\left(2x-1\right)\left(2x+1\right)\)

\(=2x^3-28x^2+98x+7x^2-98x+343-2x\left(4x^2-1\right)\)

\(=2x^3-28x^2+7x^2+343-8x^3+2x\)

\(=-6x^3-21x^2+343+2x\)

c) \(\left(4x-7\right)\left(16x^2+28x+49\right)\left(3x+1\right)\left(9x^2-3x+1\right)-9x\left(3x^2-1\right)\)

\(=\left(64x^3-343\right)\left(3x+1\right)\left(9x^2-3x+1\right)-27x^3+9x\)

\(=\left(6x^3-343\right)\left(27x^3+1\right)-27x^3+9x\)

\(=1728x^6+64x^3-9261x^3-343-27x^3+9x\)

\(=1728x^6-9224x^3-343+9x\)

27 tháng 7 2017

\(b,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\) \(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)\(\Leftrightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)

\(c,\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6x^2+12x+6-49=0\)\(\Leftrightarrow24x=-13\Rightarrow x=-\dfrac{13}{24}\)

\(d,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow-2x=23\Rightarrow x=-\dfrac{23}{2}\)

27 tháng 7 2017

T.Thùy Ninh sai câu d nha \(-2x+8=15\)

\(-2x=15-8\)

\(-2x=7\)

\(x=\dfrac{-7}{2}=-3,5\)

18 tháng 7 2016

a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)

\(3x=5\)

\(x=\frac{5}{3}\)

b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)

\(3x-8=2x-7\)

\(x=1\)

c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)

\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)

\(4x^2-3x-18=4x^2+3x\)

\(6x=-18\)

\(x=-3\)

d) Sai đề

e) ko bt

18 tháng 6 2018

a)\(16x^2-\left(4x-5\right)^2=15\)

\(\Rightarrow\left(4x\right)^2-\left(4x-5\right)^2-15=0\)

\(\Rightarrow\left(4x-4x+5\right)\left(4x+4x-5\right)-15=0\)

\(\Rightarrow5\left(8x-5\right)-15=0\)

\(\Rightarrow40x-25-15=0\)

\(\Rightarrow40x-40=0\)

\(\Rightarrow x=1\)

18 tháng 6 2018

b)\(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Rightarrow\left(4x^2-12x-9\right)-4\left(x^2-1\right)-49=0\)

\(\Rightarrow4x^2-12x+9-4x^2+4-49=0\)

\(\Rightarrow12x-36=0\)

\(\Rightarrow12\left(x-3\right)=0\)

\(\Rightarrow x=3\)

7 tháng 2 2020

\(a,\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=2\\4x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

Vậy ............

\(b,\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2,3x=6,9\\0,1x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)

Vậy ...........

\(c,\left(4x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=-2\\x^2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-0,5\\x\in\varnothing\end{cases}}\)

Vậy .........................

\(d,\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x+7=0\\x-5=0\\5x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=-7\\x=5\\5x=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{7}{2}\\x=5\\x=-\frac{1}{5}\end{cases}}\)

Vậy ...............

7 tháng 2 2020

a) \(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)

c) \(\left(4x+2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\)

\(\Leftrightarrow x=-\frac{1}{2}\) ( do \(x^2+1\ge1>0\forall x\) )

d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\x-5=0\end{cases}hoặc5x+1=0}\)

\(\Leftrightarrow x\in\left\{-\frac{7}{2},5,-\frac{1}{5}\right\}\)