e) \(\left(x+3\right)^3=\left(2x\right)^3\)

\(\Rightarrow x+3=2x\)

\(\Rightarrow2x-x=3\)

\(\Rightarrow x=3\)

f) \(\left(5-x\right)^5=32\)

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

\(\Rightarrow5-x=2\)

\(\Rightarrow x=5-2\)

\(\Rightarrow x=3\)

g) \(\left(5x-6\right)^3=64\)

\(\Rightarrow\left(5x-6\right)^3=4^3\)

\(\Rightarrow5x-6=4\)

\(\Rightarrow5x=4+6\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=\dfrac{10}{2}\)

\(\Rightarrow x=5\)

h) \(5\cdot9^x=405\)

\(\Rightarrow9^x=\dfrac{405}{5}\)

\(\Rightarrow9^x=81\)

\(\Rightarrow9^x=9^2\)

\(\Rightarrow x=2\)

i) \(11^5:11^{n-2}=11^5\)

\(\Rightarrow11^{n-2}=11^5:11^5\)

\(\Rightarrow11^{n-2}=1\)

\(\Rightarrow11^{n-2}=11^0\)

\(\Rightarrow n-2=0\)

\(\Rightarrow n=2\)

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

\(\Rightarrow3x=2x+1\)

\(\Rightarrow3x-2x=1\)

\(\Rightarrow x=1\)

b: \(B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\)

\(=2x^2-6x-2x^2+4x+2x-4\)

=-4

\(a.x=-0,6\)

\(c.x=-11,6\)

Pt nhju ak!!!

a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)

\(\Leftrightarrow x\cdot\left(x-2\right)-2x\left(x+5\right)=0\)

\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)

\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vậy: S={0;-8}

b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)

c) Ta có: \(x^2+6x+9=4x^2\)

\(\Leftrightarrow\left(x+3\right)^2-\left(2x\right)^2=0\)

\(\Leftrightarrow\left(x+3-2x\right)\left(x+3+2x\right)=0\)

\(\Leftrightarrow\left(-x+3\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: S={3;-1}

d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)

25/7nha

a, ĐK: \(x\ge11\)

\(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)

\(\Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{x^2-x+11}=16\)

\(\Leftrightarrow2x+2\sqrt{x^2-x+11}=16\)

\(\Leftrightarrow x+\sqrt{x^2-x+11}=8\)

Ta thấy \(x+\sqrt{x^2-x+11}>11>\text{​​}8\)

\(\Rightarrow\) phương trình vô nghiệm.

\(a,\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\left(x\ge11\right)\\ \Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{\left(x+\sqrt{x-11}\right)\left(x-\sqrt{x-11}\right)}=16\\ \Leftrightarrow2x+2\sqrt{x^2-x+11}=16\\ \Leftrightarrow x+\sqrt{x^2-x+11}=8\\ \Leftrightarrow\sqrt{x^2-x+11}=8-x\\ \Leftrightarrow x^2-x+11=x^2-16x+64\\ \Leftrightarrow15x=53\\ \Leftrightarrow x=\dfrac{53}{15}\left(ktm\right)\)

\(b,\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\\ \Leftrightarrow\sqrt{2x-5}+3+\left|\sqrt{2x-5}-1\right|=4\\ \Leftrightarrow\left|\sqrt{2x-5}-1\right|=1-\sqrt{2x-5}\\ \Leftrightarrow\sqrt{2x-5}-1\le0\\ \Leftrightarrow\sqrt{2x-5}\le1\\ \Leftrightarrow2x-5\le1\Leftrightarrow x\le\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{2}\)

b, ĐK: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)

\(\Leftrightarrow\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}-1\right|=4\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\):

\(\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}-1\right|\)

\(=\left|\sqrt{2x-5}+3\right|+\left|1-\sqrt{2x-5}\right|\)

\(\ge\left|\sqrt{2x-5}+3+1-\sqrt{2x-5}\right|\)

\(=4\)

Đẳng thức xảy ra khi: 

\(\left(\sqrt{2x-5}+3\right)\left(1-\sqrt{2x-5}\right)\ge0\)

\(\Leftrightarrow1-\sqrt{2x-5}\ge0\)

\(\Leftrightarrow\sqrt{2x-5}\le1\)

\(\Leftrightarrow0\le2x-5\le1\)

\(\Leftrightarrow\dfrac{5}{2}\le x\le3\)

a: \(\Leftrightarrow x\left(2x+10\right)-x\left(x-2\right)=0\)

=>x(2x+10-x+2)=0

=>x(x+12)=0

=>x=0 hoặc x=-12

b: \(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

=>(2x-5)(3x+12)=0

=>x=5/2 hoặc x=-4

c: \(\Leftrightarrow\left(2x\right)^2-\left(x+3\right)^2=0\)

=>(x-3)(3x+3)=0

=>x=3 hoặc x=-1

d: \(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

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

=>(x+2)(-5x+3)=0

=>x=-2 hoặc x=3/5

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

\(\Leftrightarrow\left(x-2\right)x-2x\left(x+5\right)=0\)

\(\Leftrightarrow x.\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-12\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-12\end{matrix}\right.\)

tìm x nhé!

1.|x+1|+|3−2x|=|3x−2|

mà |3−2x|=|3x−2| nên |x+1|=0 => x+1=0 =>x=-1

b: 2x+5=x-5

=>2x-x=-5-5

=>x=-10

c: 2x(x+2)+5(x-2)=0

=>\(2x^2+4x+5x-10=0\)

=>\(2x^2+9x-10=0\)

\(\text{Δ}=9^2-4\cdot2\cdot\left(-10\right)=81+80=161>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-9-\sqrt{161}}{4}\\x_2=\dfrac{-9+\sqrt{161}}{4}\end{matrix}\right.\)

h: 

ĐKXĐ: \(x\notin\left\{2;-1\right\}\)

\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

=>\(\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

=>\(2\left(x-2\right)-\left(x+1\right)=3x-11\)

=>2x-4-x-1=3x-11

=>x-5=3x-11

=>x-3x=-11+5

=>-2x=-6

=>x=3(nhận)

i: 3x-12=0

=>3x=12

=>x=12/3=4

f: \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)

=>\(\dfrac{3\left(x-3\right)+5\left(2x+1\right)}{15}=6\)

=>\(\dfrac{3x-9+10x+5}{15}=6\)

=>13x-4=90

=>13x=94

=>\(x=\dfrac{94}{13}\)

1) 3x-(4+2x)=11

    3x - 4 - 2x = 11

   3x - 2x = 11+4

     x = 15

Vậy x = 15.

2) 4x+5=11+2x

    4x - 2x = 11 - 5

    2x = 6

    x   = 6 : 2

     x  = 3

Vậy x  = 3.

3) 7x-17=31+x

    7x - x = 31  + 17

      6x = 48

        x = 48 : 6

       x  = 8

Vậy x = 8

# HOK TỐT #

\(3x-\left(4+2x\right)=11\)

\(3x-4-2x=11\)

\(x-15=0\)

\(x=15\)

\(4x+5=11+2x\)

\(2x-6=0\)

\(2x=6\)

\(x=3\)

Bn lm phần c đi , cố lên ! 

1) 3x-(4+2x)=11

    3x-4-2x=11

    3x-2x=11+4

          x=15

2) 4x+5=11+2x

    4x-2x=11-5

         2x=6

           x=6:2

           x=3

3) 7x-17=31+x

    7x-x=31+17

        6x=48

          x=48:6

          x=8