2/3, 3/5,4/7,5/9

thanks chúc bạn học tốt

Gọi số tự nhiên cần tìm là x.(ĐK x > 0 , x thuộc N)

Theo đề bài ta có : \(\frac{37-x}{50}=\frac{5}{9}\)

=> \(37-x=\frac{5}{9}\cdot50=\frac{250}{9}\)

=> \(x=37-\frac{250}{9}=\frac{83}{9}\)

Mà đk đã cho x thuộc N => x không thỏa mãn

Vậy không có giá trị x để thỏa mãn điều kiện cần tìm

hình như đề sai hay sao ý giải hoài ko ra :(

a) \(\sqrt{1-x^2}\) có nghĩa

\(\Leftrightarrow1-x^2\ge0\)

\(\Leftrightarrow\left(1-x\right)\left(x+1\right)\ge0\)

\(\Leftrightarrow-1\le x\le1\)

b) \(\sqrt{\frac{1}{\left(x-5\right)^2}}\)có nghĩa

\(\Leftrightarrow\frac{1}{\left(x-5\right)^2}>0\)

\(\Leftrightarrow x\ne5\)

Vậy .............

a) Để \(\sqrt{1-x^2}\)có nghĩa 

    \(\Rightarrow\)\(1-x^2\ge0\)

  \(\Leftrightarrow\)\(\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)\ge0\)

   Vì \(\sqrt{x}\ge0\forall x\)\(\Rightarrow\)\(\sqrt{x}+1\ge1>0\forall x\)

   mà \(\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)\ge0\)

   \(\Rightarrow\)\(1-\sqrt{x}\ge0\)

  \(\Leftrightarrow\)\(\sqrt{x}\le1\)

  \(\Leftrightarrow\)\(x\le1\)

Vậy để \(\sqrt{1-x^2}\)có nghĩa thì \(x\le1\)

b) Để \(\sqrt{\frac{1}{\left(x-5\right)^2}}\)có nghĩa

    \(\Rightarrow\)\(\sqrt{\frac{1}{\left(x-5\right)^2}}\ge0\)

   \(\Leftrightarrow\)\(\frac{1}{\left|x-5\right|}\ge0\)

   Vì \(1>0\)mà \(\frac{1}{\left|x-5\right|}\ge0\)

   \(\Rightarrow\)\(\left|x-5\right|>0\)( vì là mẫu số )

  \(\Leftrightarrow\)\(x-5>0\)

  \(\Leftrightarrow\)\(x>5\)

 Vậy để \(\sqrt{\frac{1}{\left(x-5\right)^2}}\)có nghĩa thì \(x>5\)

1.các p/s đó là:

1/24;24/1;2/12;12/2;3/8;8/3;4/6;6/4.

2.

a,1212/1515=4/5

b,363636/545454=2/3

k cho mình nha!

thanks

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

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

Dấu "=" xảy ra \(\Leftrightarrow\left(x+2\right)\left(2-x\right)\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2\ge0\\2-x\ge0\end{cases}}\) hoặc \(\orbr{\begin{cases}x+2\le0\\2-x\le0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\ge-2\\x\le2\end{cases}}\) hoặc \(\orbr{\begin{cases}x\le-2\\x\ge2\end{cases}}\left(vo-ly\right)\)

Vậy minQ = 4 \(\Leftrightarrow-2\le x\le2\)

Bài 1 :

ĐKXĐ : \(x\ge2\)

\(2x+5=6\sqrt{2x-4}\)

\(\Leftrightarrow4x^2+20x+25=36\left(2x-4\right)\)

\(\Leftrightarrow4x^2+20x+25-72x+144=0\)

\(\Leftrightarrow4x^2-52x+159=0\)

Đến đây chịu :))

a) \(\sqrt{25x^2-10x+1}=x+2\)

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

<=> \(\left|5x-1\right|=x+2\)

TH1: 5x - 1 \(\ge\)0 <=> x \(\ge\)1/5

Khi đó pt trở thành: 5x - 1 = x + 2

<=> 4x = 3 <=> x = 3/4 (tm)

TH2: 5x - 1 < 0 <=>  x < 1/5

Khi đó pt trở thành:  1 - 5x = x + 2

<=> -6x = 1 <=> x = -1/6 (tm)

Vậy S = {3/4; -1/6}

b) \(\sqrt{4x^2+12x+9}=7\)

<=> \(\sqrt{\left(2x+3\right)^2}=7\)

<=> \(\left|2x+3\right|=7\)

TH1: 2x + 3 \(\ge\)0 <=> x \(\ge\)-3/2

Khi đó pt trở thành: 2x + 3 = 7 <=> 2x = 4 <=> x = 2 (Tm)

TH2: 2x + 3 < 0 <=> x < -3/2

Khi đó pt trở thành: -2x - 3 = 7

<=> -2x = 10 <=> x = -5 (tm)

Vậy S = {-5; 2}

Có:     \(sin^2\phi=\frac{1}{1+cot^2\phi}=\frac{1}{a^2+1}\),  Từ đây ta được các đẳng thức:

 \(sin2\phi=2sin\phi cos\phi=2cot\phi sin^2\phi=\frac{2a}{a^2+1}\)

\(cos2\phi=1-2sin^2\phi=1-\frac{2}{a^2+1}=\frac{a^2-1}{a^2+1}\)

Xét: \(sin\left(2\phi-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(sin2\phi-cos2\phi\right)=\frac{\sqrt{2}}{2}\left(\frac{2a}{a^2+1}-\frac{a^2-1}{a^2+1}\right)=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}\left(a+1\right)}{a^2+1}\)

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

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

\(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}=\sqrt{\frac{3}{7}}\)

\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)

\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)

\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)

\(C=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

\(C=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(C=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)

\(C=\sqrt{6+2\sqrt{3}-2}\)

\(C=\sqrt{4+2\sqrt{3}}\)

\(C=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

1) Ta có: \(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)

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

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

         \(=\sqrt{2}+1+\sqrt{2}-1\)

         \(=2\sqrt{2}\approx2,82843\)

2) Ta có: \(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)

        \(\Leftrightarrow B=\frac{\sqrt{5}.\sqrt{3}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\)

        \(\Leftrightarrow B=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}\)

        \(\Leftrightarrow B=\frac{\sqrt{3}}{\sqrt{7}}\approx0,65465\)

3) Ta có: \(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)

        \(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\)

        \(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{8}.\sqrt{3-\sqrt{3}-1}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{2.8-2.2.\sqrt{3}.2}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{4.3}.2+1}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{12}.2+4}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{\left(\sqrt{12}-2\right)^2}}\)

        \(\Leftrightarrow C=\sqrt{6+\sqrt{12}-2}\)

        \(\Leftrightarrow C=\sqrt{3+2\sqrt{3}+1}\)

        \(\Leftrightarrow C=\sqrt{\left(\sqrt{3}+1\right)^2}\)

        \(\Leftrightarrow C=\sqrt{3}+1\approx2,73205\)

bạn ơi, chữ c là gì vậy, không hiểu

a)ĐKXĐ : \(x\ge1\)

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

b)Đặt  \(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

\(\Leftrightarrow A^2=2-\sqrt{3}+2+\sqrt{3}+2\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)\)

\(\Leftrightarrow A^2=4+2=6\)

\(\Leftrightarrow A=\pm\sqrt{6}\)