\(2\cdot11^x=\left(3^2+2\right)^3:\left(5^3-2^5:2^3\right)\)

\(\Leftrightarrow11^x\cdot2=1331:121\)

\(\Leftrightarrow11^x\cdot2=11\)

=> Phương trình vô nghiệm

a: =>7(x-5)>0

=>x-5>0

=>x>5

b: =>x-1 thuộc {1;-1;11;-11}

=>x thuộc {2;0;12;-10}

c: =>x+1+7 chia hết cho x+1

=>x+1 thuộc {1;-1;7;-7}

=>x thuộc {0;-2;6;-8}

d: =>(x+2)(x-5)<0

=>-2<x<5

a:(- 7) . ( 5 – x) < 0

=>7(x-5)>0

=>x-5>0

=>x>5

b:11 ⁝ x – 1

=>x-1 thuộc {1;-1;11;-11}

=>x thuộc {2;0;12;-10}

c: x + 8 ⁝ x + 1

=>x+1+7 chia hết cho x+1

=>x+1 thuộc {1;-1;7;-7}

=>x thuộc {0;-2;6;-8}

d: (x + 2) . (5 – x) > 0

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

=>-2<x<5

\(\left(x-1\right)^2=5^2\\\Rightarrow x-1=5\\ \Rightarrow x=5+1=6\)

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

b) \(=x^2-2.x.10+10^2=\left(x-10\right)^2\)

c) \(=x\left(x+3\right)+2\left(x+3\right)=\left(x+3\right)\left(x+2\right)\)

\(\dfrac{3}{x+2}\)=\(\dfrac{5}{2x+1}\)(x khác -1/2 và x khác -2)

=>6x+3=5x+10

<=>x=7 tm 

Ta có \(f\left(1\right)=5\)

\(\Leftrightarrow2\cdot a\cdot1-1=5\)

\(\Leftrightarrow2a=6\)

\(\Leftrightarrow a=3\)

Vậy đa thức có dạng \(f\left(x\right)=2ax-1=6x-1\)

Ta có \(f\left(3\right)=6\cdot3-1=17\)

Vậy \(f\left(3\right)=17\)

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

\(\Rightarrow-x=\frac{5}{4}+2=\frac{13}{4}\)

\(\Rightarrow x=\frac{-13}{4}\)

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

\(-x=\frac{5}{4}+2\)

\(-x=\frac{13}{4}\)

\(x=-\frac{13}{4}\)

-x - 2 = 5/4

=> -x = 5/4 + 2

=> -x = 13/4

=> x = -13/4

\(\frac{2}{5}-\frac{1}{2}\left(x+\frac{1}{3}\right)=\frac{7}{5}\)

\(\frac{1}{2}\left(x+\frac{1}{3}\right)=\frac{2}{5}-\frac{7}{5}\)

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

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

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

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

\(x=-\frac{7}{3}\)

\(\frac{2}{5}-\frac{1}{2}.\left(x+\frac{1}{3}\right)=\frac{7}{5}\)

\(\Rightarrow\frac{1}{2}.\left(x+\frac{1}{3}\right)=\frac{2}{5}-\frac{7}{5}\)

\(\Rightarrow\frac{1}{2}.\left(x+\frac{1}{3}\right)=-1\)

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

\(\Rightarrow x=-\frac{7}{3}\)

115x ² + x ² -x + 1/4 + 15/4 = (x-1/2) ² +115x ² + 15/4 ≥ 15/4

 ⇒ pt vô nghiệm