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

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

a)Ta có :

  3x chia hết cho x+1

=>3x+3-3 chia hết cho x+1

=>-3 chia hết cho x+1

=>x+1 thuộc Ư(-3)

=>x+1 thuộc {1;-1;3;-3}

=>x thuộc {0;-2;2;-4}

b)Ta có :

 5x+2 chia hết cho x+1

=>5x+5-3 chia hết cho x+1

=>-3 chia hết cho x+1

=>x+1 thuộc Ư(-3)

=>x+1 thuộc {1;-1;3;-3}

=>x thuộc {0;-2;2;-4}

mai cậu đi học à

 \(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=2\)

\(\le\left|x\right|-\left|2\right|+\left|x\right|-\left|3\right|+\left|x\right|-\left|4\right|\)

\(x-2+x-3+x-4=2\)

\(\Leftrightarrow x+x+x-2-3-4=2\)

\(\Leftrightarrow x^3-2-3-4=2\)

\(x^3=2+4+3+2\)

\(x^3=11\)

\(x=\sqrt[3]{11}\)

e/(x+6)(x-1)(x²+5x+16)

Help me!!!

câu a, b tách dấu trị tuyệt đối ra thôi. 
câu c pt đc thành (x+1)(2x^2+....)=0=>...
 

Em nghĩ là như vầy ạ:

\(B=\frac{4-x+x+1}{\left(4-x\right)\left(x+1\right)}=\frac{5}{-x^2+3x+4}\) (-1 < x < 4)

Ta có: \(-x^2+3x+4=-\left(x-\frac{3}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Do đó: \(B=\frac{5}{-x^2+3x+4}\ge\frac{5}{\frac{25}{4}}=\frac{20}{25}=\frac{4}{5}\)

Vậy min B = 4/5 khi x = 3/2 (TMĐK)

1/(x + 1) + 1/(4 - x) ≥ (1 + 1)^2/(x + 1 + 4 - x) = 4/5

Có: \(\frac{1}{x\left(x+1\right)}\)= \(\frac{1}{x}-\frac{1}{x+1}\)

Mà \(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2017}\)

=> \(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2017}\)

=> \(-\frac{1}{x+1}\)= \(\frac{1}{x}+\frac{1}{2017}-\frac{1}{x}\)

=> \(-\frac{1}{x+1}=\frac{1}{2017}\)

=> \(-1\cdot2017=\left(x+1\right)\cdot1\)

=> \(-2017=x+1\)

=> \(x=-2017-1\)

=> \(x=-2018\)

Vậy \(x=-2018\)

x = -2018

Sorrry, có cái dòng đấu viết nhầm 2017

Ta có: \(3^x+3^{x+1}+3^{x+2}=351\)

\(\Rightarrow3^x.1+3^x.3+3^x.3^2^{ }=351\)

\(\Rightarrow3^x.1+3^x.3+3^x.9=351\)

\(\Rightarrow3^x.\left(1+3+9\right)=351\)

\(\Rightarrow3^x.13=351\)

\(\Rightarrow3^x=351:13=27\)

\(\Rightarrow x=3\)

3 k mk nha

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3^x+3^x+1+3^x+2=351

3^x+3^x.3+3^x.3²=351

3^x.(1+3+3²)=351

3^x.(1+3+9)=351

3^x.13=351

3^x=351:13

3^x=27

3^x=3³

  x=3

Vậy x=3