Hỏi đáp Căn thức

Ta có

  \(\frac{7}{5}=\frac{\sqrt{49}}{5}\)

     \(\sqrt{2}=\frac{5\sqrt{2}}{5}=\frac{\sqrt{50}}{5}\)

   Vì \(\frac{\sqrt{50}}{5}>\frac{\sqrt{49}}{5}\Rightarrow\sqrt{2}>\frac{7}{5}\)

còn \(\sqrt{2}< \frac{19}{20}\) là vô lí đấy . Vì

  \(\sqrt{2}>1;\frac{19}{20}< 1\)

\(\sqrt{2}=\frac{5\sqrt{2}}{5}=\frac{\sqrt{50}}{5}>\frac{\sqrt{49}}{5}=7\)

\(\sqrt{2}=\frac{20\sqrt{2}}{20}=\frac{\sqrt{800}}{20}>\frac{\sqrt{361}}{20}=\frac{19}{20}\)

Ta có\(\sqrt{x+5}\ge0\Rightarrow\sqrt{x+5}-3\ge-3\Rightarrow Bmin=-3\)

\(\sqrt{x^2-9}\ge0\Rightarrow\sqrt{x^2-9}+5\ge5\Rightarrow Cmin=5\)

Làm j có đề bài đâu mà lm

         Bài làm :

Ta có :

\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\frac{1}{2-\sqrt{3}}\)

\(=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\frac{2+\sqrt{3}}{4-3}\)

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

\(=\sqrt{3}-2-\sqrt{3}\)

\(=-2\)

\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\frac{1}{2-\sqrt{3}}\)

\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\frac{\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=\sqrt{3}-2-\sqrt{2}=-2\)

dòng cuối là \(\sqrt{3}-2-\sqrt{3}=-2\)nhá

Check lại đề đi bạn ơi! Chứng minh \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\) thì được

Đặt \(D=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Leftrightarrow D^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)

\(\Leftrightarrow D^2=8+2\sqrt{16-10-2\sqrt{5}}\)

\(\Leftrightarrow D^2=8+2\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow D^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(\Leftrightarrow D^2=8+2\left(\sqrt{5}-1\right)\)

\(\Leftrightarrow D^2=6+2\sqrt{5}\)

\(\Leftrightarrow D^2=\left(\sqrt{5}+1\right)^2\)

\(\Rightarrow D=\sqrt{5}+1\)

Thay vào ta tính được: \(A=\sqrt{5}+1-\sqrt{5}=1\)

Vậy A = 1

\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\)    ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))

\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)

\(=\sqrt{4\cdot\sqrt{7}}\)

\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)

\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)

\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)

\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)

\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)

\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)

\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)

\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)

\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)

\(\Leftrightarrow4\sqrt{21}\)

cuối lười tính nên thôi nhá :>

tks :>

a) Ta có: \(\frac{x-\sqrt{x}}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\sqrt{x}}{\sqrt{x}-1}\)

=> đk: \(x\ge0;x\ne1\)

b) đk: \(x\ge0\)

c) đk: \(x\ge-2\)

a) \(\sqrt{\left(2x-6\right)^2}=\left|2x-6\right|=2x-6\)

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

\(\sqrt{\left(2x-6\right)^2}=\left|2x-6\right|=2x-6\left(x\ge3\right)\)

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

a) Ta có: \(\sqrt{\left(2x-6\right)^2}\)

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

\(=2\left(x-3\right)=2x-6\) (vì \(x\ge3\))

b) Ta có: \(\sqrt{\left(x-4\right)^2}\)

\(=4-x\) (vì x<4)

Bài làm:

đk: \(x\ge0\)

Ta có: \(A=\frac{x-\sqrt{x}-5}{\sqrt{x}+3}=\frac{\left(x-\sqrt{x}-12\right)+7}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)+7}{\sqrt{x}+3}\)

\(=\sqrt{x}-4+\frac{7}{\sqrt{x}+3}\)

+ Nếu: x không là số chính phương => \(\sqrt{x}\) vô tỉ

=> A vô tỉ (loại)

+ Nếu: x là số chính phương

=> \(\sqrt{x}\) nguyên

Khi đó để A nguyên => \(\frac{7}{\sqrt{x}+3}\inℤ\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(7\right)\)

Mà \(\sqrt{x}+3\ge3\left(\forall x\right)\) => \(\sqrt{x}+3=7\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\)

Vậy khi x = 16 thì biểu thức A đạt giá trị nguyên

thanks FL.Shizuka nha !

\(\left(\frac{4}{3}\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)

\(=\frac{4\sqrt{10}}{5}+\frac{4\sqrt{6}}{3}-\frac{16\sqrt{15}}{15}+\frac{2\sqrt{15}}{5}+4-\frac{4\sqrt{10}}{5}+2+\frac{2\sqrt{15}}{3}-\frac{4\sqrt{6}}{3}\)

\(=4\)

bạn phải tính từng bước chứ @

Mấy cái giống nhau trừ đi . Đằng nào chả ra 4 . Tự làm bước  đó được mà ??

d) Tại \(x=13-4\sqrt{10}\)

\(P=3+\frac{2}{\sqrt{13-4\sqrt{10}}+2}\)

\(P=3+\frac{2\sqrt{2}}{\sqrt{26-8\sqrt{10}}+2\sqrt{2}}\)

\(P=3+\frac{2\sqrt{2}}{\sqrt{16-8\sqrt{10}+10}+2\sqrt{2}}\)

\(P=3+\frac{2\sqrt{2}}{\sqrt{\left(4-\sqrt{10}\right)^2}+2\sqrt{2}}\)

\(P=3+\frac{2\sqrt{2}}{4-\sqrt{10}+2\sqrt{2}}\)

\(P=\frac{12-3\sqrt{10}+6\sqrt{2}+2\sqrt{2}}{4-\sqrt{10}+2\sqrt{2}}\)

\(P=\frac{12-3\sqrt{10}+8\sqrt{2}}{4-\sqrt{10}+2\sqrt{2}}\)

a) \(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\) \(\left(x\ge0;x\ne1\right)\)

\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\frac{7}{2}\)

\(\Leftrightarrow\frac{3\sqrt{x}+8}{\sqrt{x}+2}=\frac{7}{2}\)

\(\Rightarrow6\sqrt{x}+16=7\sqrt{x}+14\)

\(\Leftrightarrow\sqrt{x}=2\Rightarrow x=4\)

\(\left(\sqrt{6}-2\sqrt{3}+5\sqrt{2}-\frac{1}{2\sqrt{8}}\right).2\sqrt{6}\)

\(=2.6-12\sqrt{3}+20\sqrt{3}-\frac{\sqrt{3}}{2}\)

\(=\frac{24+15\sqrt{3}}{2}\)

a) \(ĐKXĐ:x>0;x\ne4\)

Ta có : \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4x}{2\sqrt{x}-x}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}\right)\)

\(=\left[\frac{\sqrt{x}.\sqrt{x}-4x}{\sqrt{x}.\left(\sqrt{x}-2\right)}\right]\cdot\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

\(=\frac{-3x}{\sqrt{x}.\left(\sqrt{x}+3\right)}\)

b) Ta có : \(x-1=10-4\sqrt{6}=\left(\sqrt{6}-2\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{6}-2\right)^2+1}\)

......

\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right).\sqrt{2}.\left(\sqrt{5}-\sqrt{1}\right).\sqrt{3-\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right).\sqrt{6-2\sqrt{5}}\)

\(=\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right).\sqrt{5-2\sqrt{5}+1}\)

\(=\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\)

\(=\frac{6+2\sqrt{5}}{2}.\left(\sqrt{5}-1\right)^2\)

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

\(=\frac{\left(5-1\right)^2}{2}=\frac{4^2}{2}=8\)

 \(x\left(5x^3+2\right)-2\left(\sqrt{2x+1}-1\right)=0\)

\(\Leftrightarrow5x^4+2x-2\sqrt{2x+1}+2=0\)

\(\Leftrightarrow5x^4+\left(2x+1-2\sqrt{2x+1}+1\right)=0\)

\(\Leftrightarrow5x^4+\left(\sqrt{2x+1}-1\right)^2=0\)

có \(\hept{\begin{cases}5x^4\ge0\\\left(\sqrt{2x+1}-1\right)^2\ge0\end{cases}}\)mà \(5x^4+\left(\sqrt{2x+1}-1\right)^2=0\Rightarrow\hept{\begin{cases}5x^4=0\\\left(\sqrt{2x+1}-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^4=0\\\sqrt{2x+1}=1\end{cases}\Leftrightarrow x=0}\)

vạy x=0 là nghiệm của phương trình

Cre: Đàm Hải Ngọc

cái này dùng liên hợp dễ hơn 

\(x\left(5x^3+2\right)-2\left(\sqrt{2x+1}-1\right)=0\left(đk:x\ge-\frac{1}{2}\right)\)

\(< =>x\left(5x^3+2\right)-2.\frac{2x+1-1}{\sqrt{2x+1}+1}=0\)

\(< =>x\left(5x^3+2\right)-x.\frac{4}{\sqrt{2x+1}+1}=0\)

\(< =>x\left(5x^3+2-\frac{4}{\sqrt{2x+1}+1}\right)=0< =>x=0\)

giờ dùng đk đánh giá cái ngoặc to vô nghiệm là ok

mk giải giúp bn cho, một mk mk giải thôi

căn thức tối giản :>>

Đặt \(A=2\sqrt{3}\left(2\sqrt{6}-\sqrt{3}+1\right)\)

\(A=4\sqrt{18}-2.3+2\sqrt{3}\)

\(A=12\sqrt{2}+2\sqrt{3}-6\)