3/5 x = 5/6

x = 5/6 : 3/5

x = 25/18

\(\dfrac{3}{5}:x=\dfrac{1}{3}+\dfrac{1}{2}\)

\(\dfrac{3}{5}:x=\dfrac{5}{6}\)

\(x=\dfrac{3}{5}:\dfrac{5}{6}\)

\(x=\dfrac{18}{25}\)

3/5:*=1/3+1/2

3/5:*=5/6

*=5/6:3/5

*=25/18

Bài 1:

\(x^3-x^2-x+1=0\)

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

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

\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy x = 1 hoặc x = -1

Bài 2:
\(2x-2x^2-1=-2\left(x^2-x+\dfrac{1}{2}\right)\)

\(=-2\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}\right)\)

\(=-2\left(x^2-\dfrac{1}{2}\right)^2-\dfrac{1}{2}< 0\)

\(\Rightarrowđpcm\)

  ( x + 1 )² +3( x - 5)(x+ 5)-( 2x-1)²

=x² + 2x + 1 + 3(x² - 25) - 4x² - 4x + 1

= x² + 2x + 1 + 3x² - 75 - 4x² - 4x + 1

= -2x - 73

k cho mk nhe!!

( x + 1 )² +3( x - 5)(x+ 5)-( 2x-1)²

=x²+2x+1+3x²-75-4x²+4x-1

=(x²+3x²-4x²)+(2x+4x)-(1-1)-75

=6x-75

Vậy ms đúng bn kia sai r`

bạn coi lại đề

ai giup tra loi voi

2 bai tim x do

\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)

\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)

\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)

\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)

\(=-\left(2x+\frac{7}{5}\right)^m\)

đến đây thì mình chịu

a. 3^x=1-x^2

x=0 la nghiem

x>=1;  VT>=3 VP<=0 vo nghiem 

b. (de bai thieu n khac 0 vi neu n=0 dung voi moi x)

3x-14=1=> x=5

c.(5^2x5^^x+1)=5^4

5^{^x+1}=5^2=> x=1

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

\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)

\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)

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

\(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)

\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)

\(\frac{5}{3}-\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{7}{6}\)

=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)

=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{2}\)

=> \(\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{3}:\frac{1}{2}=\frac{1}{3}\cdot2=\frac{2}{3}\)

=> \(1-\frac{x}{3}=\frac{2}{3}\)

=> \(\frac{x}{3}=1-\frac{2}{3}=\frac{1}{3}\)

=> x = 1

\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)-\frac{1}{4}=\frac{1}{2}\)

=> \(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)

=> \(x:\frac{1}{2}+\frac{3}{2}=3-\frac{3}{4}=\frac{9}{4}\)

=> \(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)

=> \(x:\frac{1}{2}=\frac{3}{4}\)

=> \(x=\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}\)

\(\frac{5}{3}-\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{7}{6}\)

\(\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)

\(\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{1}{2}\)

\(1-x\times\frac{1}{3}=\frac{1}{3}:\frac{1}{2}\)

\(1-x\times\frac{1}{3}=\frac{2}{3}\)

\(x\times\frac{1}{3}=1-\frac{2}{3}\)

\(x\times\frac{1}{3}=\frac{1}{3}\)

\(x=\frac{1}{3}:\frac{1}{3}\)

\(x=1\)

\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)-\frac{1}{4}=\frac{1}{2}\)

\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{1}{2}+\frac{1}{4}\)

\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{3}{4}\)

\(x:\frac{1}{2}+\frac{3}{2}=3-\frac{3}{4}\)

\(x:\frac{1}{2}+\frac{3}{2}=\frac{9}{4}\)

\(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)

\(x:\frac{1}{2}=\frac{3}{4}\)

\(x=\frac{3}{4}\times\frac{1}{2}\)

\(x=\frac{3}{8}\)