Ta có \(x^4+2x^2+1=\left(x^2+1\right)^2\)

          Ta thấy \(\left(x^2+1\right)^2>0\forall x\)

\(\Rightarrow\)đa thức trên không có nghiệm

Vậy ...

`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) Q=\(\left(\dfrac{2x+1}{\sqrt{x}^3-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x}^3}{1+\sqrt{x}}-\sqrt{x}\right)\)

=\(\left(\dfrac{2x+1-x+\sqrt{x}}{\sqrt{x}^3-1}\right)\left(\dfrac{1+\sqrt{x}^3-\sqrt{x}-x}{1+\sqrt{x}}\right)\)

=\(\dfrac{\sqrt{x}+x+1}{\sqrt{x}^3-1}.\left(-2\sqrt{x}+1\right)\)

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

b) ta có : Q=3 => \(\dfrac{-2\sqrt{x}+1}{\sqrt{x}-1}=3=>-2\sqrt{x}+1=3\sqrt{x}-3\)

=>x=16/25=0,64

vậy x=0,64 khi Q=3

a) Q=\(\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)

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

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

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

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

b) ta có : Q=3 <=> \(\sqrt{x}-1=3\)

\(\Leftrightarrow\) \(\sqrt{x}=4\Leftrightarrow x=16\)

vậy để Q=3 thì x=16

cái này ms đúng

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=>...
 

a) Ta có : |2x - 5| = x + 1

\(\Leftrightarrow\orbr{\begin{cases}2x-5=-x-1\\2x-5=x+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+x=-1+5\\2x-x=1+5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=4\\x=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=6\end{cases}}\)

mik ko pc

Vì \(\left(2x+1\right).\left(y-3\right)=10\)nên 2x + 1 và y - 3 thuộc ước của 10 

Mà \(Ư\left(10\right)=\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

Ta thấy 2x +1 là số lẻ nên \(2x+1\in\left\{1;-1;5;-5\right\}\)

Ta có bảng sau 

Vậy.....

ta có \(\left(2x+1\right)\left(y-3\right)=10.1=1.10\)

Nếu \(2x+1=10\Rightarrow x=\frac{9}{2}\)( ko tồn tại vì \(x\in Z\))

       \(y-3=1\Rightarrow y=4\)

Nếu \(2x+1=1\Rightarrow x=0\)

          \(y-3=10\Rightarrow y=13\)

vậy cặp số nguyên  \(x,y\)cần tìm là: \(x;y\left(0;13\right)\)

Vì 2x + 1 . y − 3 = 10 nên 2x + 1 và y - 3 thuộc ước của 10
Mà Ư 10 = 1; − 1;2; − 2;5; − 5;10; − 10
Ta thấy 2x +1 là số lẻ nên 2x + 1 ∈ 1; − 1;5; − 5
Kết luận : ..............

a) \(\sqrt{x-2}+\sqrt{16x-32}=10\)

\(\Rightarrow\sqrt{x-2}+4\sqrt{x-2}=10\)

\(\Rightarrow5\sqrt{x-2}=10\)

\(\Rightarrow\sqrt{x-2}=2\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

b) \(\sqrt{x+\sqrt{2x-1}}=5\sqrt{2}\)

ĐK \(x\ge\dfrac{1}{2}\)

\(\sqrt{x+\sqrt{2x-1}}=5\sqrt{2}\)

\(\left(\sqrt{x+\sqrt{2x-1}}\right)^2=\left(5\sqrt{2}\right)^2\)

\(\left|x+\sqrt{2x-1}\right|=50\)

\(\sqrt{2x-1}=50-x\)

\(\left(\sqrt{2x-1}\right)^2=\left(50-x\right)^2\)

\(\left|2x-1\right|=x^2-100x+2500\)

\(2x-1=x^2-100x+2500\)

\(x=41\)

a) ĐK x\(\ge\)2

\(\sqrt{x-2}+\sqrt{16x-32}=10\)

\(\sqrt{x-2}+\sqrt{16\left(x-2\right)}=10\)

\(\sqrt{x-2}+4\sqrt{x-2}=10\)

\(\left(1+4\right)\sqrt{x-2}=10\)

\(5\sqrt{x-2}=10\)

\(\sqrt{x-2}=2\)

\(\left(\sqrt{x-2}\right)^2=4\)

\(\left|x-2\right|=4\)(vì \(x\ge2\))

\(x-2=4\)

\(x=6\)