b: \(\left(2x+1\right)^2=25\)

=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left(1-3x\right)^3=64\)

=>\(\left(1-3x\right)^3=4^3\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1

d: \(\left(4-x\right)^3=-27\)

=>\(\left(4-x\right)^3=\left(-3\right)^3\)

=>4-x=-3

=>x=4+3=7

e: \(x^2-5x=0\)

=>\(x\left(x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

|x+1|+|3x-1|+|x-1|=3

=} vs cả trong dấu giá trị tuyệt đối >0 thì=}

x+1+3x-1+x-1=3{=}5x=4{=}x=4/5

=}vs cả trong giá trị tuyệt đối <0 thì=}

x+1+3x-1+x-1=-3{=}5x=-4{=}x=-4/5

a: \(\Leftrightarrow x^2-x+4x-4+5⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{2;0;6;-4\right\}\)

c: \(\Leftrightarrow4x-5=13k\left(k\in Z\right)\)

=>4k=13k+5

hay \(x=\dfrac{13k+5}{4}\)

`123-x=38`

`=> x= 123-38`

`=>x=85`

Vậy `x=85`

__

`(x+15) -7=33`

`=>x+15=33+7`

`=>x+16= 40`

`=>x=40-16`

`=>x=24`

Vậy `x=24`

__

`3^(x+3) -13=230`

`=> 3^(x+3) = 230+13`

`=>3^(x+3)=243`

`=> 3^(x+3)=3^5`

`=> x+3=5`

`=>x=5-3`

`=>x=2`

Vậy `x=2`

Bài 1:

\(a,A=2x^2+2x+1=\left(x^2+2x+1\right)+x^2=\left(x+1\right)^2+x^2\\ Mà:\left(x+1\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x+1\right)^2+x^2>0\forall x\in R\\ Vậy:A>0\forall x\in R\)

2:

a: =-(x^2-3x+1)

=-(x^2-3x+9/4-5/4)

=-(x-3/2)^2+5/4 chưa chắc <0 đâu bạn

b: =-2(x^2+3/2x+3/2)

=-2(x^2+2*x*3/4+9/16+15/16)

=-2(x+3/4)^2-15/8<0 với mọi x

Bài 1:

\(B=4+x^2+x=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\\ Vậy:B>0\forall x\in R\)

\(b,3\left(x-2\right)+2\left(3x-5\right)=10\\ \Leftrightarrow3x-6+6x-10=10\\ \Leftrightarrow3x+6x=10+10+6\\ \Leftrightarrow9x=26\\ \Leftrightarrow x=\dfrac{26}{9}\\ c,2x-\left(3x+1\right)=5x-2\\ \Leftrightarrow2x-3x-1=5x-2\\ \Leftrightarrow2x-3x-5x=-2+1\\ \Leftrightarrow-6x=-1\\ \Leftrightarrow x=\dfrac{1}{6}\\ d,3x+2=-5+6 \\ \Leftrightarrow3x=-5+6-2\\ \Leftrightarrow3x=-2\\ \Leftrightarrow x=-\dfrac{1}{3}\)

a: =>3x-6+6x-10=10

=>9x=26

=>x=26/9

b: =>5x-2=2x-3x-1

=>5x-2=-x-1

=>6x=1

=>x=1/6

d: =>3x+2=1

=>3x=-1

=>x=-1/3

b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)

=>-6x+16=0

=>-6x=-16

hay x=8/3(nhận)

c: \(\Leftrightarrow\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+2}\)

\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)

\(\Leftrightarrow2x^2+4x-2x^2+2=0\)

=>4x+2=0

hay x=-1/2(nhận)

b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)

\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)

\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)

\(\Leftrightarrow3x\left(x+4\right)=0\)

=>x=0(nhận) hoặc x=-4(loại)

a: Ta có: \(x\left(2-3x\right)+\left(3x^3-x^2\right):x\)

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

=x

b: Ta có: \(2x\left(x-3y\right)-\left(8x^3y-12x^2y^2\right):2xy\)

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

\(=-2x^2\)