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6 tháng 5 2017

a) Gợi ý: a 2 − 5 a + 4 = ( a − 1 ) ( a − 4 ) ; a 2 + 3 a − 4 = ( a − 1 ) ( a + 4 )  

Ta rút gọn được A = a + 1 a − 4  

b) Thay a = 5 vào biểu thức A tìm được A = 6

c) Ta biến đổi A = a + 1 a − 4 = 1 + 5 a − 4  

⇒ A ∈ ℤ ⇒ a ∈ − 1 ;   3 ;   5 ;   9

5 tháng 7 2021

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 \(A_{min}=-\dfrac{1}{4}\)

5 tháng 7 2021

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 \(a=\dfrac{1}{4}\)

13 tháng 4 2017

a) \(A=\left(\dfrac{3}{4}:\dfrac{2}{3}\right):\dfrac{3}{5}=\left(\dfrac{3}{4}\cdot\dfrac{3}{2}\right)\cdot\dfrac{5}{3}=\dfrac{9}{8}\cdot\dfrac{5}{3}=\dfrac{15}{8}\)

b) \(B=\dfrac{3}{4}:\left(\dfrac{2}{5}:\dfrac{3}{5}\right)=\dfrac{3}{4}:\left(\dfrac{2}{5}\cdot\dfrac{5}{3}\right)=\dfrac{3}{4}:\dfrac{2}{3}=\dfrac{3}{4}\cdot\dfrac{3}{2}=\dfrac{9}{8}\)

c) \(C=\left(\dfrac{11}{12}:\dfrac{33}{16}\right)\cdot\dfrac{3}{5}=\left(\dfrac{11}{12}\cdot\dfrac{16}{33}\right)\cdot\dfrac{3}{5}=\dfrac{4}{9}\cdot\dfrac{3}{5}=\dfrac{4}{15}\)

13 tháng 4 2017

a.

A=\(\left(\dfrac{3}{4}:\dfrac{2}{3}\right):\dfrac{3}{5}\)

=\(\left(\dfrac{3}{4}.\dfrac{3}{2}\right):\dfrac{3}{5}\)

=\(\dfrac{9}{8}:\dfrac{3}{5}\)

=\(\dfrac{9}{8}.\dfrac{5}{3}=\dfrac{45}{24}=\dfrac{15}{8}\)

b.

B=\(\dfrac{3}{4}:\left(\dfrac{2}{5}:\dfrac{3}{5}\right)\)

=\(\dfrac{3}{4}:\left(\dfrac{2}{5}.\dfrac{5}{3}\right)\)

=\(\dfrac{3}{4}:\dfrac{2}{3}=\dfrac{3}{4}.\dfrac{3}{2}\)

=\(\dfrac{9}{8}\)

c.

C=\(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}\)

=\(\left(\dfrac{11}{12}.\dfrac{16}{33}\right).\dfrac{3}{5}\)

=\(\dfrac{4}{9}.\dfrac{3}{5}=\dfrac{12}{45}\)

=\(\dfrac{4}{15}\)

25 tháng 9 2021

\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

27 tháng 10 2020

a) đk: \(\hept{\begin{cases}a\ge0\\a\ne16\end{cases}}\)

Ta có: 

\(C=\frac{a}{a-16}-\frac{2}{\sqrt{a}-4}-\frac{2}{\sqrt{a}+4}\)

\(C=\frac{a-2\cdot\left(\sqrt{a}+4\right)-2\cdot\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)

\(C=\frac{a-2\sqrt{a}-8-2\sqrt{a}+8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)

\(C=\frac{a-4\sqrt{a}}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}=\frac{\sqrt{a}}{\sqrt{a}+4}\)

b) Ta có: \(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)

\(\Rightarrow\sqrt{a}=\sqrt{5}-2\)

Khi đó: \(C=\frac{\sqrt{5}-2}{\sqrt{5}-2+4}=\frac{\sqrt{5}-2}{\sqrt{5}+2}=\frac{\left(\sqrt{5}-2\right)^2}{1}=9-4\sqrt{5}\)

27 tháng 10 2020

\(C=\frac{a}{a-16}-\frac{2}{\sqrt{a}-4}-\frac{2}{\sqrt{a}+4}\)

a) ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne16\end{cases}}\)

\(=\frac{a}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}-\frac{2\left(\sqrt{a}+4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}-\frac{2\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)

\(=\frac{a-2\sqrt{a}-8-2\sqrt{a}+8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)

\(=\frac{a-4\sqrt{a}}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)

\(=\frac{\sqrt{a}\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}=\frac{\sqrt{a}}{\sqrt{a}+4}\)

b) Với \(a=9-4\sqrt{5}\)( tmđk )

\(C=\frac{\sqrt{a}}{\sqrt{a}+4}=1-\frac{4}{\sqrt{a}+4}\)

\(C=1-\frac{4}{\sqrt{9-4\sqrt{5}}+4}\)

\(=1-\frac{4}{\sqrt{5-4\sqrt{5}+4}+4}\)

\(=1-\frac{4}{\sqrt{\left(\sqrt{5}-2\right)^2}+4}\)

\(=1-\frac{4}{\left|\sqrt{5}-2\right|+4}\)

\(=1-\frac{4}{\sqrt{5}-2+4}\)

\(=1-\frac{4}{\sqrt{5}+2}\)

\(=\frac{\sqrt{5}+2-4}{\sqrt{5}+2}\)

\(=\frac{\sqrt{5}-2}{\sqrt{5}+2}\)

\(=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}-2\right)}{1}=9-4\sqrt{5}\)

17 tháng 6 2016

(a+b+c)3=(a+b)3+3(a+b)2c+3(a+b)c2+c3

=a3+b3+3ab.(a+b)+3(a+b)2c+3(a+b)c2+c3

=a3+b3+c3+3(a+b)(ab+ac+bc+c2)

=a3+b3+c3+3(a+b)[a.(b+c)+c.(b+c)]

=a3+b3+c3+3(a+b)(b+c)(c+a) 

=>dpcm

17 tháng 6 2016

 

P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=>2P=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=(52-1)(52+1)(54+1)(58+1)(516+1)

=(54-1)(54+1)(58+1)(516+1)

=(58-1)(58+1)(516+1)

=(516-1)(516+1)

=532-1

==>P=(532-1)/2

 

 

25 tháng 12 2020

a ⇒A=\(4\sqrt{4\times3}+3\sqrt{25\times3}-5\sqrt{16\times3}=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}=3\sqrt{3}\)

b ĐKXĐ x≥2 ⇔\(\sqrt{x-2}+3\sqrt{x-2}=16\Leftrightarrow4\sqrt{x-2}=16\Leftrightarrow\sqrt{x-2}=4\Rightarrow x-2=16\Leftrightarrow x=18\)

25 tháng 12 2020

a.   \(A=4\sqrt{12}+3\sqrt{75}-5\sqrt{48}\)

          \(=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}\)

          \(=3\sqrt{3}\)

b.    \(\sqrt{x-2}-\sqrt{9x-18}=16\)

   \(\Leftrightarrow\sqrt{x-2}-\sqrt{9\left(x-2\right)}=16\)

   \(\Leftrightarrow\sqrt{x-2}-3\sqrt{x-2}=16\)

   \(\Leftrightarrow-2\sqrt{x-2}=16\)

   \(\Leftrightarrow\sqrt{x-2}=-8\)  ( Vô lý )

   Vậy PT vô nghiệm