ta có: 5/x = 1/8 => x= 1/8 x 5 = 5/8

=> x= 8

duyệt đi

Ta có : \(x^2+x+4=x^2+x+\frac{1}{4}+\frac{15}{4}=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\left(\forall x\right)\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2+x=-4\end{cases}}\)

+) x² + x = - 4

<=> ( x + 1/2 )² = - 4 + 1/4 = -15/4

Mà ( x + 1/2 )² lớn hơn hoặc bằng 0 với mọi x

=> x² + x + 4 = 0 ktm

Vậy pt = 0 <=> x = 1

ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ge0\\x-2>0\\x+2>0\\x\ge0\end{matrix}\right.\)  và \(4-x\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x>2\\x>-2\\x\ge0\end{matrix}\right.\) và \(x\ne4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)

a: ĐKXĐ: \(x\ne\dfrac{3}{2}\)

b: ĐKXĐ: \(x\in R\)

đề bài yêu cầu tìm x hay tìmB

tìm giá trị nhỏ nhất

bạn giải đi mình căm ơn bạn trước

A=x^3y^2+(2xy-8xy)+(-5+6)+(-x^3y)+x^2

A=x^3y^2+(-6xy)+1+(-x^3y)+x^2

Bậc là 3

B=(2xy-5xy+12xy)+(-8+11)+x^2y^2+4x^2y

B=9xy+3+x^2y^2+4x^2y

Bậc là 2;thay x=-1,y=-1 vào A ta đc

cứ thế ban làm tiếp nha

`(x+1)(x+3)=2x^2-2`

`<=>x^2+x+3x+3=2x^2-2`

`<=>x^2-4x-5=0`

`<=>x^2-5x+x-5=0`

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

`<=>(x-5)(x+1)=0`

`<=>` $\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.$

Vậy `S={5,-1}`

Ta có: \(\left(x+1\right)\left(x+3\right)=2x^2-2\)

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

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

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

\(\Leftrightarrow\left(x+1\right)\left[x+3-2\left(x-1\right)\right]=0\)

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

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

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

Vậy: S={-3;5}

a: =>x^2006-x=0

=>x(x^2005-1)=0

=>x=0 hoặc x=1

b: =>5^x*26=650

=>5^x=25

=>x=2

c: =>x(2y-1)-y+1/2=5/2

=>(y-1/2)(2x-1)=5/2

=>(2y-1)(2x-1)=5

=>\(\left(2x-1;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(0;-2\right);\left(-2;0\right)\right\}\)

\(\frac{3x-5}{2x+7}=\frac{4x+8}{5x-4}\)

\(\left(3x-5\right)\left(5x-4\right)=\left(4x+8\right)\left(2x+7\right)\)

\(15x^2-12x-10x+20=8x^2+28x+16x+56\)

\(15x^2-8x^2-12x-10x-28x-16x+20-56=0\)

\(7x^2-66x-36=0\)

bạn tự làm tiếp nha

(3x-5)/(2x+7)=(4x+8)/(5x-4)

<=>(3x-5)(5x-4)=(4x+8)(2x+7)

<=>15x-20=8x+56

<=>15x-8x=56+20

<=>7x=76

<=>x=76:7=76/7

\(\frac{3x-5}{2x+7}=\frac{4x+8}{5x-4}\)

\(\left(3x-5\right)\left(5x-4\right)=\left(4x+8\right)\left(2x+7\right)\)

\(3x.5x-4.3x-5.5x+5.4=4x.2x+7.4x+8.2x+8.7\)

\(15x^2-12x-10x+20=8x^2+28x+16x+56\)

\(15x^2-22x+20=8x^2+44x+56\)

\(15x^2-22x+20-8x^2-44x-56=0\)

\(7x^2-66x-36=0\)