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1:
a: \(\sqrt{36}-\sqrt{100}=6-10=-4\)
b: Để \(\sqrt{\dfrac{2}{2x-1}}\) có nghĩa thì \(\dfrac{2}{2x-1}>=0\)
=>2x-1>0
=>x>1/2
2:
a: \(A=\dfrac{\left(15\sqrt{180}-5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}\)
\(=15\sqrt{\dfrac{180}{10}}-5\sqrt{\dfrac{200}{10}}-3\sqrt{\dfrac{450}{10}}\)
\(=15\sqrt{18}-5\sqrt{20}-3\sqrt{45}\)
\(=45\sqrt{2}-10\sqrt{5}-9\sqrt{5}\)
\(=45\sqrt{2}-19\sqrt{5}\)
b: \(B=\sqrt{32}-\sqrt{50}-16\sqrt{\dfrac{1}{8}}\)
\(=4\sqrt{2}-5\sqrt{2}-\dfrac{16}{\sqrt{8}}\)
\(=-\sqrt{2}-2\sqrt{8}=-\sqrt{2}-4\sqrt{2}=-5\sqrt{2}\)
a)\(\left(\frac{\sqrt{8}}{x-1}\right)^2=\left(\sqrt{2}\right)^2\Leftrightarrow\frac{8}{x^2-2x+1}=2\Leftrightarrow\frac{8}{x^2-2x+1}-2=0\)
\(\Rightarrow\frac{8-2.\left(x^2-2x+1\right)}{x^2-2x+1}=0\Rightarrow8-2x^2-2x-2=0\Rightarrow-2x^2+4x+6=0\)
\(\Rightarrow-2x^2+6x-2x+6=0\Rightarrow-2x\left(x+1\right)+6\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(6-2x\right)\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Câu sau tương tự nếu ko biết thì nhắn tin cho mình nha chọn cho mình nha cảm ơn
Bài 1:
a: \(\sqrt{27}+\dfrac{1}{2}\sqrt{48}-\sqrt{108}\)
\(=3\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-6\sqrt{3}\)
\(=-3\sqrt{3}+2\sqrt{3}=-\sqrt{3}\)
b: \(\left(\sqrt{14}-\sqrt{10}\right)\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{2}\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{12+2\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)=7-5=2\)
c: \(\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)
\(=\sqrt{3}-\sqrt{3}-1=-1\)
Bài 2:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5+\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: A=2
=>\(\sqrt{x}=2\left(\sqrt{x}-1\right)\)
=>\(2\sqrt{x}-2=\sqrt{x}\)
=>\(\sqrt{x}=2\)
=>x=4(nhận)
c: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1+1⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\inƯ\left(1\right)\)
=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)
=>\(\sqrt{x}\in\left\{2;0\right\}\)
=>\(x\in\left\{4;0\right\}\)
Bạn chỉ cần lam cho trong căn xuất hiện hằng đẵng thức là được
VD:\(\sqrt{2+2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\left(\sqrt{2}+1\right)\)
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~
a, \(=\sqrt{\left(2\sqrt{2}\right)^2+2\times2\sqrt{2}\times\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}=2\sqrt{2}+\sqrt{5}\)
\(a,\)
\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :
\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)
\(\Leftrightarrow-3\sqrt{x}+11=0\)
\(\Leftrightarrow-3\sqrt{x}=-11\)
\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)
\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{121}{9}\)
Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)
\(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)}=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}=2\sqrt{5}\)
1)
a) \(\sqrt{2x-4}\) có nghĩa khi:
\(2x-4\ge0\)
\(\Leftrightarrow2x\ge4\)
\(\Leftrightarrow x\ge\dfrac{4}{2}\)
\(\Leftrightarrow x\ge2\)
b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi
\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)
\(\Rightarrow4-x\le0\)
\(\Leftrightarrow x\ge4\)
Cô hướng dẫn nhé :)
1. ĐK: \(x\ge0\)
\(pt\Leftrightarrow x+5=x+1+2\sqrt{x}\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\left(tm\right)\)
2. \(A=\sqrt{10}+\sqrt{6}.\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{2}\)
1) bình phương 2 vế là ra
2) A=\(\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{2}\cdot2=2\sqrt{2}\)