help me!!!
Rút gọn: \(\sqrt{\left(\sqrt{3}+1\right)2^{ }}-\sqrt{4-2\sqrt{3}}\)
Giải PT: \(\dfrac{3}{5}\). \(\sqrt{25x-50}\) - \(\sqrt{x-2}\) = 6
Rút gọn:
1) \(\dfrac{16-6\sqrt{7}}{\sqrt{7}-3}\)
2) \(\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
3) \(\dfrac{\left(\sqrt{3}+2\sqrt{5}\right)^2-8\sqrt{15}}{\sqrt{6}-2\sqrt{10}}\)
Giúp em với ạ. Help mee !!!
Câu 1,2 bạn đã đăng và có lời giải rồi
Câu 3:
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)
Rút gọn biểu thức sau
A=\(\dfrac{1}{x-1}\sqrt{75\left(x-1\right)^3}\left(x>1\right)
\)
B=\(5\sqrt{4x}-3\sqrt{\dfrac{100x}{9}}-\dfrac{4}{x}\sqrt{\dfrac{x^3}{4}}\left(x>0\right)
\)
C=\(x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)
Help me
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
Trục căn thức ở mẫu và rút gọn
a)\(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
b)\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
Help me plssssssss
a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
\(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{3^2-\left(\sqrt{5}\right)^2}}\)
\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)
\(=\dfrac{3-\sqrt{5}}{2}\)
b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
\(=\dfrac{\left|2-\sqrt{3}\right|}{\sqrt{4-3}}\)
\(=\dfrac{2-\sqrt{3}}{1}\)
\(=2-\sqrt{3}\)
a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)
b: \(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{1}}=2-\sqrt{3}\)
d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
thực hiện phép tính
\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)
giải phương trình
\(\sqrt{x-3}=6\)
\(\sqrt{\left(x-3\right)^2}=12\)
rút gọn biểu thức
a) \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right).\left(\dfrac{3-\sqrt{x}}{3-x}\right)\) (với x≥0 ; x≠3; x≠9
b) \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right)\div\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\) (x >0)
c) \(A=\sqrt{3x-1}+3.\sqrt{12x-4}-\sqrt{6^2.\left(3x-1\right)}+\sqrt{5}\) với x≥ \(\dfrac{1}{3}\)
d) \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\) với a>0,a≠1, a≠ \(\pm\)2
Bài 1:
\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=4-\sqrt{5}+\sqrt{5}+1=5\)
Bài 2:
a: ĐKXĐ: x>=3
\(\sqrt{x-3}=6\)
=>x-3=36
=>x=36+3=39(nhận)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x-3\right)^2}=12\)
=>\(\left|x-3\right|=12\)
=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Bài 3:
a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)
\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)
\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)
\(=\dfrac{-\sqrt{x}\left(x-3\right)-\left(x-3\right)}{-\left(x-3\right)}=\dfrac{\left(x-3\right)\left(\sqrt{x}+1\right)}{x-3}=\sqrt{x}+1\)
b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)
\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)
\(=\sqrt{3x-1}+\sqrt{5}\)
d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\left(a-2\right)}{a+2}\)
Trục căn thức và rút gọn
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
Help me plssssssss
\(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}\right)^2-1^2}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}\right)^2-2^2}-\dfrac{12\left(3+\sqrt{6}\right)}{3^2-\left(\sqrt{6}\right)^2}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}\right)^2-11^2\)
\(=6-121\)
\(=-115\)
giải pt
a.\(\sqrt{\left(x-3\right)^2}=2\)
b.\(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
a) \(\sqrt{\left(x-3\right)^2}=2\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\left(đk:x\ge-2\right)\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\Leftrightarrow\sqrt{x+2}=3\Leftrightarrow x+2=9\Leftrightarrow x=7\)
a: \(\sqrt{\left(x-3\right)^2}=2\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
* giải phương trình
a. \(\sqrt{\left(x-3\right)^2}=2\)
b.\(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
a) Ta có: \(\sqrt{\left(x-3\right)^2}=2\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) ĐKXĐ: \(x\ge-2\)
Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+\dfrac{4}{5}\cdot5\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7(thỏa ĐK)
a) \(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy:.....
b) ĐKXĐ: x ≥ -2
\(\Leftrightarrow\sqrt{9}.\sqrt{x+2}-5.\sqrt{x+2}+\dfrac{4}{5}.\sqrt{25}.\sqrt{x+2}=6\)
<=> \(\sqrt{x+2}.\left(3-5+\dfrac{4}{5}.5\right)=6\)
\(\Leftrightarrow2.\sqrt{x+2}=6\)
\(\Leftrightarrow\sqrt{x+2}=3\)
<=> x + 2 = 9
<=> x = 7
Tham khảo ạ:
a) Ta có: √(x−3)2=2(x−3)2=2
⇔|x−3|=2⇔|x−3|=2
⇔[x−3=2x−3=−2⇔[x=5x=1⇔[x−3=2x−3=−2⇔[x=5x=1
b) ĐKXĐ: x≥−2x≥−2
Ta có: ⇔3√x+2−5√x+2+45⋅5√x+2=6⇔3x+2−5x+2+45⋅5x+2=6
⇔2√x+2=6⇔2x+2=6
⇔x+2=9⇔x+2=9
hay x=7(thỏa ĐK)
* Giải phương trình
a. \(\sqrt{\left(x-3\right)^2}=2\)
b. \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
* Cho Q= \(\dfrac{1}{x-2\sqrt{x}+3}\)
Tìm giá trị lớn nhất của Q
a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)
\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)
\(Q=\dfrac{1}{x-2\sqrt{x}+3}\)
Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)
Rút gọn các biểu thức sau
a)\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
b)\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
c)\(x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)
Help me plsss
\(a,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\\ =\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\=2\sqrt{2} \)
\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)
\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\) (vì \(x>4\) )
@seven