Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐK: \(x\ge-1\)
pt <=> \(\left(14\sqrt{x+35}-84\right)+\left(6\sqrt{x+1}-\sqrt{x^2+36x+35}\right)=0\)
<=> \(14\left(\sqrt{x+35}-6\right)+\sqrt{x+1}\left(6-\sqrt{x+35}\right)=0\)
<=> \(\left(\sqrt{x+35}-6\right)\left(11-\sqrt{x+1}\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x+35}-6=0\\11-\sqrt{x+1}=0\end{cases}}\)Em làm tiếp nhé!
Cau 1:
a: \(A=\dfrac{\left(\sqrt{a}-2\right)\left(a+2\sqrt{a}+4\right)+2\sqrt{a}\left(\sqrt{a}-2\right)}{a-4}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(a+4\sqrt{a}+4\right)}{a-4}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}=\sqrt{a}+2\)
c: \(=\dfrac{\left|c+1\right|}{\left|c\right|-1}\)
TH1: c>0
\(C=\dfrac{c+1}{c-1}\)
TH2: c<0
\(C=\dfrac{\left|c+1\right|}{-\left(c+1\right)}=\pm1\)
Áp dụng BĐT:\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
Ta có: \(\left|\sqrt{x-1}+2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+2+3-\sqrt{x-1}\right|=5\)
Dấu \(=\)xảy ra khi \(AB\ge0\)
dat \(\sqrt{x-1}\) = t
ta có: \(\sqrt{x+3+4t}\)+ \(\sqrt{x+8-6t}\)= 5
x + 3 + 4t + x + 8 - 6t = 25
2x - 2t = 14 ( chia cả 2 vế cho 2)
x - t = 7
t = x - 7
thay t = \(\sqrt{x}-1\)vào ta được:
x - 7 = \(\sqrt{x-1}\)
( x - 7 )2 = x - 1
x2 -14x + 49 = x - 1
x2 - 15x + 50 = 0
k biết đúng hay k
a: ĐKXĐ: x>=0
b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)
\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)
\(\Leftrightarrow\sqrt{4x\left(\sqrt{x}+2\right)}=4\sqrt{2}-\sqrt{2x}\)
\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)
\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)
=>\(x\in\left\{0;1.2996\right\}\)
Em làm bừa thôi, mới học dạng này .
ĐK: \(1\le x\le7\)
Đặt \(\sqrt{6}\ge a=\sqrt{7-x}\ge0;\sqrt{6}\ge b=\sqrt{x-1}\ge0\)
PT<=>\(b^2+2a=2b+ab\left(1\right)\)
(1) \(\Leftrightarrow\left(a-b\right)\left(2-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\b=2\end{cases}}\). Nếu a = b thì \(\sqrt{7-x}=\sqrt{x-1}\Leftrightarrow7-x=x-1\Leftrightarrow x=4\) (TM)
Nếu b = 2 thì \(\sqrt{x-1}=2\Leftrightarrow x=5\left(TM\right)\)
Vậy...
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 \(\sqrt{x}⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1+1⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\inƯ\left(1\right)\)
=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)
=>\(\sqrt{x}\in\left\{2;0\right\}\)
=>\(x\in\left\{4;0\right\}\)