1) x - 2 = -6

x = -6 + 2

x = -4

2) -5 . x - ( -3 ) =13

-5 . x = 13 + ( -3 )

-5 . x = 10

x = 10 : ( -5 )

x = -2

\(\left(a+b\right)\times c=a\times c+b\times c\)

\(a\times\left(b+c\right)=a\times b+a\times c\)

\(a\times\left(b-c\right)=a\times b-a\times c\)

\(a\times b-a\times c=a\times\left(b-c\right)\)

( a+b)xc------axc+bxc

a x ( b + c)----- a x b + a x c

a x ( b-c )------ a x b - a x c

a x b-a x c-------a x ( b - c )

đây: 2xb-b tức là 2xb-1xb đúng ko?

vì 12 tháng cộng 12 tháng bằng 2 năm

\(ĐKXĐ:x\ne\pm1\)

Ta có : \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)

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

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

\(\Leftrightarrow\frac{2}{\left(x^2+x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x+2\right)^2}{\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{2\left(x+1\right)\left(x-1\right)-2\left(x+2\right)^2}{\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow2\left(x^2-1\right)-2\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow2x^2-2-2x^2-8x-8=0\)

\(\Leftrightarrow-8x-10=0\)

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

Vậy \(x=-\frac{5}{4}\) là nghiệm của phương trình.

a/ \(f\left(x\right)\ge2\sqrt{\frac{16x^2}{x^2}}=8\)

Dấu "=" xảy ra khi \(x^2=\frac{16}{x^2}\Leftrightarrow x=\pm2\)

b/ Hàm này không tồn tại GTNN

c/ \(f\left(x\right)=x+3+\frac{25}{x+3}-4\ge2\sqrt{\frac{25\left(x+3\right)}{x+3}}-4=6\)

Dấu "=" xảy ra khi \(x+3=\frac{25}{x+3}\Leftrightarrow x=2\)

d/ \(f\left(x\right)=x+\frac{9}{x}+3\ge2\sqrt{\frac{9x}{x}}+3=9\)

Dấu "=" xảy ra khi \(x=\frac{9}{x}\Leftrightarrow x=3\)

bn này bị j rùi mọi người!đây mà là toán lớp 3!

T   =   [ x 2 + ( a − b ) x − a b x 2 − ( a − b ) x − a b . x 2 − ( a + b ) x + a b x 2 + ( a + b ) x + a b ] : [ x 2 − ( b − 1 ) x − b x 2 + ( b + 1 ) x + b . x 2 − ( b + 1 ) x + b x 2 − ( 1 − b ) x − b ] =   [ ( x − b ) ( x + a ) ( x − a ) ( x + b ) . ( x − a ) ( x − b ) ( x + a ) ( x + b ) ] : [ ( x − b ) ( x + 1 ) ( x + b ) ( x + 1 ) . ( x − 1 ) ( x − b ) ( x + b ) ( x − 1 ) ] =   ( x − b ) 2 ( x + b ) 2 : ( x − b ) 2 ( x + b ) 2 = 1

Vậy T = 1

Đáp án cần chọn là: A

1: A={x∈R|x<=2}

=>A=(-∞;2]

B={x∈R|x>5}

=>B=(5;+∞)

A\(\cap\) B=(-∞;2]\(\cap\) (5;+∞)

=∅

A\(\cup\) B=(-∞;2]\(\cup\) (5;+∞)

A\B=(-∞;2]\(5;+∞)

=(-∞;2]

B\A=(5;+∞)\(-∞;2]

=(5;+∞)

\(C_{R}A\) =R\A=R\(-∞;2]

=(2;+∞)

\(C_{R}B=\) R\B=R\(5;+∞)

=(-∞;5]

2: A={x∈R|x<0 hoặc x>=2}

=>A=(-∞;0)\(\cup\) [2;+∞)

B={x∈R|-4<=x<3}

=>B=[-4;3)

A\(\cap\) B=((-∞;0)\(\cup\) [2;+∞))\(\cap\) [-4;3)

=[-4;0)\(\cup\) [2;3)

A\(\cup\) B=((-∞;0)\(\cup\) [2;+∞))\(\cup\) [-4;3)

=(-∞;+∞)

A\B=((-∞;0)\(\cup\) [2;+∞))\[-4;3)

=(-∞;-4)\(\cup\) [3;+∞)

B\A=[-4;3)\((-∞;0)\(\cup\) [2;+∞))

=[0;2)

\(C_{R}A\) =R\A=[0;2)

\(C_{R}B\) =R\B=R\[-4;3)

=(-∞;-4)\(\cup\) [3;+∞)

3: |x-1|<2

=>-2<x-1<2

=>-1<x<3

=>A=(-1;3)

|x+1|<3

=>-3<x+1<3

=>-4<x<2

=>B=(-4;2)

A\(\cap\) B=(-1;3)\(\cap\) (-4;2)

=(-1;2)

A\(\cup\) B=(-1;3)\(\cup\) (-4;2)

=(-4;3)

A\B=(-1;3)\(-4;2)

=[2;3)

B\A=(-4;2)\(-1;3)

=(-4;-1]

\(C_{R}A\) =R\A=R\(-1;3)

=(-∞;-1]\(\cup\) [3;+∞)

\(C_{R}B\) =R\B=R\(-4;2)

=(-∞;-4]\(\cup\) [2;+∞)