\(=A\cdot\left(40+60+80-180\right)+2023\)

=0+2023

=2023

=>x+20=0 hoặc x+4=0

=>x=-4 hoặc x=-20

`(x+20)^100 . |x+4|=0`

TH1: `(x+20)^100 =0`

`x+20=0`

`x=-20`

TH2: `|x+4|=0`

`x+4=0`

`x=-4`

Vậy `x=-20;x=-4`

\(\left(x+20\right)^{100}.\left|x+4\right|=0\)

\(< =>\left[{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|x+4\right|=0\end{matrix}\right.< =>\left[{}\begin{matrix}x+20=0\\x+4=0\end{matrix}\right.\)

\(< =>\left[{}\begin{matrix}x=0-20\\x=0-4\end{matrix}\right.< =>\left[{}\begin{matrix}x=-20\\x=-4\end{matrix}\right.\)

Vậy .............

minh moi hoc lop 5 thoi

1) 3x - 6=  5x + 2

5x - 3x = -6 - 2

2x = -8

x = -4

2) 15 - x = 4x - 5

4x + x = 15 + 5

5x = 20

x = 4

Tương tự như trên 

1) 3x - 6 = 5x + 2

  3x - 5x = 2 + 6

        -2x = 8

           x = 8 : (-2)

           x = -4

2) 15 - x = 4x - 5

    -x - 4x = -5 - 15

         -5x = -20

            x = -20 : (-5)

            x = 4

`a)`

`A=(x+1)(2x-1)`

`=2x^{2}+x-1`

`=2(x^{2}+(1)/(2)x-(1)/(2))`

`=2(x^{2}+(1)/(2)x+(1)/(16)-(9)/(16))`

`=2(x+(1)/(4))^{2}-(9)/(8)>= -9/8` với mọi `x`

Dấu `=` xảy ra khi :

`x+(1)/(4)=0<=>x=-1/4`

Vậy `min=-9/8<=>x=-1/4`

``

`b)`

`(4x+1)(2x-5)`

`=8x^{2}-18x-5`

`=8(x^{2}-(9)/(4)x-(5)/(8))`

`=8(x^{2}-(9)/(4)x+(81)/(64)-(121)/(64))`

`=8(x-(9)/(8))^{2}-(121)/(8)>= -(121)/(8)` với mọi `x`

Dấu `=` xảy ra khi :

`x-(9)/(8)=0<=>x=9/8`

Vậy `min=-121/8<=>x=9/8`

\(A=2x^2+x-1=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

\(A_{min}=-\dfrac{9}{8}\) khi \(x=-\dfrac{1}{4}\)

\(B=8x^2-18x-5=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)

\(B_{min}=-\dfrac{121}{8}\) khi \(x=\dfrac{9}{8}\)

a) \(A=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{9}{8}=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\)Vì \(2\left(x+\dfrac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{1}{4}\)

b) \(B=\left(4x+1\right)\left(2x-5\right)=8x^2-18x-5=8\left(x^2-\dfrac{9}{4}x+\dfrac{81}{64}\right)-\dfrac{121}{8}=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\)

Vì \(8\left(x-\dfrac{9}{8}\right)^2\ge0\Rightarrow8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)

\(ĐTXR\Leftrightarrow x=\dfrac{9}{8}\)

5x-16=40+x

=> 5x-16-x = 40

=> 5x-x -16=40

4x-16=40

4x= 40+16

4x=56

x= 56:4

x=14

Vậy...

4x-10=15-x

=> 4x-10+x= 15

4x+x -10=15

5x= 15+10

5x= 25

x= 25:5

x=5

Vậy....

5x -16=40+x

=> 5x-x=40+16

=>4x=56

=>x=56:4

x=14

\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\)

\(=\left(\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\right).\left(1-4x\right)\)

\(=\left(\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\right)\left(1-4x\right)\)

\(=\dfrac{-4\sqrt{x}.\left(4x-1\right)}{4x-1}=-4\sqrt{x}\)

\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\left(dkxd:x\ge0;x\ne\dfrac{1}{4}\right)\)

\(=\left[\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\right]\cdot\left(1-4x\right)\)

\(=\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\cdot\left[-\left(4x-1\right)\right]\)

\(=4\sqrt{x}\cdot\left(-1\right)\)

\(=-4\sqrt{x}\)