a) xx = x

=> xx - x = 0

x.(x-1) = 0

=> x = 0

x - 1 = 0 => x = 1

KL:...

b) Để \(\frac{1}{4x}\) là số nguyên

\(\Rightarrow1⋮4x\Rightarrow4x\inƯ_{\left(1\right)}=\left\{\pm1\right\}\)

nếu 4x = 1 => x = 1/4

4x = -1 => x = -1/4

KL:...

a) xx = x

=> xx - x = 0

x.( x - 1 ) = 0

=> x = 0

x - 1 = 0 => x = 1

KL : x = 1

b) Để \(\frac{1}{4x}\) là số nguyên.

\(\Rightarrow\)1 : 4x \(\Rightarrow\)\(\in\)Ư_{( 1)} = { \(\mp\)1 }

Nếu 4x = 1 => x = - \(\frac{1}{4}\)

KL :...

CHÚC BN HỌC GIỎI NHÉ.

\(x.x=x\)

\(x=x:x\)

\(x=1\)

\(\sqrt{-3x^3+5x+14}+\sqrt{-5x^3+6x+28}=\left(4-2x-x^2\right)\sqrt{2-x}\) (ĐKXĐ: \(x\in R,x\le2\))

\(\Leftrightarrow\sqrt{\left(2-x\right)\left(3x^2+6x+7\right)}+\sqrt{\left(2-x\right)\left(5x^2+10x+14\right)}-\left(4-2x-x^2\right)\sqrt{2-x}=0\)

\(\Leftrightarrow\sqrt{2-x}\left(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}-4+2x+x^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\left(1\right)\end{cases}}\)

Pt \(\left(1\right)\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-\left(x+1\right)^2+5\left(2\right)\)

Ta có: \(\left(x+1\right)^2\ge0\Rightarrow\sqrt{2\left(x+1\right)^2+4}\ge\sqrt{4}=2\)

Tương tự: \(\sqrt{5\left(x+1\right)^2+9}\ge3\). Từ đó: \(VT_{\left(2\right)}\)\(\ge2+3=5\)

Mà \(VP_{\left(2\right)}=-\left(x+1\right)^2+5\le5\) nên dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)(tm)

Vậy tập nghiệm của pt cho là \(S=\left\{2;-1\right\}.\)

\(M\left(x\right)=-3x^2+6x-4+2x^2-5x+4=-x^2+x\)

Đặt M(x)=0

=>-x(x-1)=0

=>x=0 hoặc x=1

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

Giả sử: \(M\left(x\right)=0\)

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

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

\(A=x^5-5x^4+5x^3-5x^2+5x-1\)

\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x+3\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x+3\)

\(=3\)

Ta có : 

\(A=x^5-5x^4+5x^3-5x^2+5x-1\)

\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x+3\)\(A=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x+3\)

\(A=3\)

P/s tham khảo nha 

hok tốt

\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=4\frac{1}{4}\)

\(\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=4\frac{1}{4}\)

\(\frac{1}{10}x=\frac{17}{4}+\frac{3}{8}\)

\(\frac{1}{10}x=\frac{37}{8}\)

\(x=\frac{185}{4}\)

\(\left(\frac{2}{3}x-\frac{1}{2}\right).\frac{3}{4}-\frac{2}{5}x=4\frac{1}{4}\)

\(\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=\frac{17}{4}\)

\(\frac{1}{2}x-\frac{2}{5}x=\frac{17}{4}+\frac{3}{8}\)

\(\frac{1}{10}x=\frac{37}{8}\)

       \(x=\frac{37}{8}:\frac{1}{10}\)

      \(x=\frac{185}{4}\)

= 3x^2-2x+2

k mình nha