b) \(\sqrt{16x}-5\left(\sqrt{x}-2\right)-\sqrt{79x}-5\)

\(=\sqrt{4^2x}-5\sqrt{x}+10-\sqrt{79x}-5\)

\(=4\sqrt{x}-5\sqrt{x}-\sqrt{79x}+5\)

\(=-\sqrt{x}-\sqrt{79x}+5\)

\(=-\sqrt{x}-\sqrt{79}.\sqrt{x}+5\)

\(=\sqrt{x}\left(-1-\sqrt{79}\right)+5\)

a) \(4\sqrt{x}-5\sqrt{4x}-\sqrt{25x}-3\sqrt{x}-5\)

\(=4\sqrt{x}-5\sqrt{2^2x}-\sqrt{5^2x}-3\sqrt{x}-5\)

\(=4\sqrt{x}-10\sqrt{x}-5\sqrt{x}-3\sqrt{x}-5\)

\(=\left(4-10-5-3\right)\sqrt{x}-5\)

\(=-14\sqrt{x}-5\)

cau b) ban viet ro de bai ra di

Lời giải:

a.

$(2x+3)^2-4(x-2)(x+2)=(4x^2+12x+9)-4(x^2-4)$

$=4x^2+12x+9-4x^2+16=12x+25$

b.

$(4x^3-16x^2+17x-5):(2x-5)$

$=[2x^2(2x-5)-3x(2x-5)+(2x-5)]:(2x-5)$

$=(2x-5)(2x^2-3x+1):(2x-5)=2x^2-3x+1$

bài 2

P= (x+1)(x²-x+1)+x-(x-1)(x²+x+1)+2010 với x = -2010

= (x³+1) + x - (x³-1) + 2010

= x³ + 1 + x - x³ + 1 + 2010

= x + 2 + 2010

= 2010 + 2 + 2010

=4022

Q=16x(4x²-5)-(4x+1)(16x²-4x + 1) với x = 1/5 

= (4x)³-16.5x - [(4x)³+1]

= (4x)³ - 16.5x - (4x)³ - 1

= -16.5x - 1

= -16.5.1/5 - 1

= -16-1

=-17

a) (x-3)(x²+3x+9)-x(x-4)(x+4)=41

<=> x³ - 3³ - x(x² - 4²) = 41

<=> x³ - 27 - x³ + 16x = 41

<=> 16x = 68

<=> x= 4,25

b) (x+2)(x²-2x+4)-x(x²+2)=4

<=> x³ + 2³ - x³  - 2x =4

<=> 8 - 2x = 4

<=> 2x = 4

<=> x= 1/2

1.X=4
2X=7

các bạn cho mình bt cách làm luôn nhé

Do x < 2 => 2 - x âm => |2 - x| = -(2 - x) = -2 + x

Ta có:

B = 5 - (-2 + x)

B = 5 + 2 - x

B = 7 - x

\(B=5-\left|2-x\right|\)

\(=5\left(2-x\right)\) (vì x<2)

\(=10-5x\)

a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

Vậy...

b)Đk: \(x\ge-1\)

Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)

Vậy...

\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)

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

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

b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)

a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)

b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)