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

2x - 2 - 5 = 15 - 9x

2x - 7 = 15 - 9x

2x + 9x = 15 + 7

11x = 22

x = 2

Vậy x = 2 

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

\(\Leftrightarrow2x-2-5=15-9x\)

\(\Leftrightarrow2x-\left(2+5\right)=15-9x\)

\(\Leftrightarrow2x-7=15-9x\)

\(\Leftrightarrow2x+9x=15+7\)

\(\Leftrightarrow11x=22\)

\(\Leftrightarrow x=22\div11\)

\(\Leftrightarrow x=2\)

\(\text{Vậy }x=2\)

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

    2x - 2 - 5 = 15 - 9x

         2x - 7 = 15 - 9x

      2x + 9x = 15 + 7

            11x = 22

                x = 2

1.: Áp dụng BĐT Cauchy-Schwarz cho 3 số dương 

\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\)

a) Ta có: \(\frac{x+a}{x+2}+\frac{x-2}{x-a}=2\left(1\right)\)

Với a = 4

Thay vào phương trình (t) ta được:

  \(\frac{x+2}{x+2}+\frac{x-2}{x-2}=2\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

\(\Leftrightarrow2x^2=2x^2-8\)

\(\Leftrightarrow0x=-8\)

Vậy phương trình vô nghiệm

b) Nếu x = -1

\(\Rightarrow\frac{-1+a}{-1+2}+\frac{-1-2}{-1-a}=2\)

\(\Leftrightarrow\frac{-1+a}{1}+\frac{-3}{-1-a}=2\)

\(\Leftrightarrow\frac{\left(-1+a\right)\left(-1-a\right)}{-1-a}+\frac{-3}{-1-a}=\frac{2\left(-1-a\right)}{-1-a}\)

\(\Leftrightarrow1+a-a-a^2-3=-2-2a\)

\(\Leftrightarrow-a^2+2a=-2-1+3\)

\(\Leftrightarrow a\left(2-a\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=0\\2-a=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)

Vậy a = {0;2}

NĂM MỚI VUI VẺ

\(a,\frac{x+4}{x+2}+\frac{x-2}{x-4}=2\)

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

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

=>\(2\left(\frac{x-4+x+2}{\left(x+2\right)\left(x-4\right)}\right)=0\)

=> x=1 (t/m \(x\ne-2\) và \(x\ne4\))

Thiếu vế phải rồi bạn

a: =>6x-2-2x<2x-1

=>4x-2<2x-1

=>2x-1<0

=>x<1/2

b: =>4x-8>=9x-6+4-2x

=>4x-8>=7x-2

=>-3x>=6

=>x<=-2

c: =>3x^2-12<3x^2+x

=>x>-12

d: =>5x^2-x+20x-4>5x^2+16x+2

=>19x-4>16x+2

=>3x>6

=>x>2

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Vậy tập nghiệm của ohuowng trình là \(S=\left\{\dfrac{1}{3};3;4\right\}\)

=>x^4+4x^2+9-4x^3-6x^2+12x<x^4-4x^3-2x^2+15x-3

=>-2x^2+12x+9<-2x^2+15x-3

=>-3x<-12

=>x>4