\(2\sqrt{x}-7=-3\sqrt{x}+13\)
\(\Leftrightarrow5\sqrt{x}=20\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy x =16
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Bài 1: Căn bậc hai
Các câu hỏi tương tự
2 tháng 4 2019 lúc 22:27
Rút gọn
\(A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{\sqrt{x}-10}{x-4}\) (x\(\ge\)0, x \(\ne\) 4)
\(B=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)
Bài 2
a) A sqrt{left(1-sqrt{2}right)^2}+sqrt{left(-2right)^6}-sqrt{left(1+sqrt{2}right)^2}
b) B sqrt{7+2sqrt{6}}+sqrt{7-2sqrt{6}}
c) C sqrt{7-4sqrt{3}}
d) D 2sqrt{7+4sqrt{3}}-sqrt{13-4sqrt{3}}
e) E frac{1}{1+sqrt{3}}+frac{1}{sqrt{3}+sqrt{5}}+...+frac{1}{sqrt{79}+sqrt{81}}
Bài 4:
a) sqrt{x-1}2
b) sqrt{x^2-3x+2}sqrt{2}
c) sqrt{4x+1}x+1
d) sqrt{x+2sqrt{x-1}}-sqrt{x-2sqrt{x-1}}2
e) sqrt{x^2-4x+5}+sqrt{x^2-4x+8}+sqrt{x^2-4x+9}3+sqrt{5}
f)
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Bài 2
a) A= \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(-2\right)^6}-\sqrt{\left(1+\sqrt{2}\right)^2}\)
b) B= \(\sqrt{7+2\sqrt{6}}+\sqrt{7-2\sqrt{6}}\)
c) C= \(\sqrt{7-4\sqrt{3}}\)
d) D= \(2\sqrt{7+4\sqrt{3}}-\sqrt{13-4\sqrt{3}}\)
e) E= \(\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{79}+\sqrt{81}}\)
Bài 4:
a) \(\sqrt{x-1}=2\)
b) \(\sqrt{x^2-3x+2}=\sqrt{2}\)
c) \(\sqrt{4x+1}=x+1\)
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
e) \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)
f)
Rút gọn các biểu thức sau:
a,sqrt{16a^2} - 5a với a ≥ 0
b, 3x + 2 - sqrt{9x^2+6x+1} với x ≥ frac{1}{3}
c,sqrt{8+2sqrt{7}} - sqrt{7}
d,sqrt{14-2sqrt{13}} + sqrt{14+2sqrt{13}}
e, 2x - sqrt{4x^2-4x+1} với x frac{1}{2}
g, |x-2| + frac{sqrt{x^2-4x+4}}{x-2} với x 2
Đọc tiếp
Rút gọn các biểu thức sau:
a,\(\sqrt{16a^2}\) - 5a với a ≥ 0
b, 3x + 2 - \(\sqrt{9x^2+6x+1}\) với x ≥ \(\frac{1}{3}\)
c,\(\sqrt{8+2\sqrt{7}}\) - \(\sqrt{7}\)
d,\(\sqrt{14-2\sqrt{13}}\) + \(\sqrt{14+2\sqrt{13}}\)
e, 2x - \(\sqrt{4x^2-4x+1}\) với x > \(\frac{1}{2}\)
g, |x-2| + \(\frac{\sqrt{x^2-4x+4}}{x-2}\) với x > 2
Bài 1:
a) sqrt{13-2sqrt{42}}
b) sqrt{46+6sqrt{5}}
c) sqrt{12-3sqrt{15}}
d) sqrt{11+sqrt{96}}
Bài 2:
a) Asqrt{6+2sqrt{2}sqrt{3-sqrt{4+2sqrt{3}}}}
b) Bsqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}
c) Csqrt{3-sqrt{5}}left(sqrt{10}+sqrt{2}right)left(3+sqrt{5}right)
d) Dsqrt{4+sqrt{7}}-sqrt{4-sqrt{7}}
e) Esqrt{8+2sqrt{10+2sqrt{5}}}+sqrt{8-2sqrt{10+2sqrt{5}}}
g) Gsqrt{x+2sqrt{x-1}}+sqrt{x-2sqrt{x-1}}
h) H4x-sqrt{9x^2-12x+4}
i) frac{sqrt{7}-sqrt{2}}{sqrt{7}+sqrt{2}}+frac{sqrt{7}+sqrt{2}}{sqrt{7}-sqrt...
Đọc tiếp
Bài 1:
a) \(\sqrt{13-2\sqrt{42}}\)
b) \(\sqrt{46+6\sqrt{5}}\)
c) \(\sqrt{12-3\sqrt{15}}\)
d) \(\sqrt{11+\sqrt{96}}\)
Bài 2:
a) \(A=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
b) \(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
c) \(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
d) \(D=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
e) \(E=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
g) \(G=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
h) \(H=4x-\sqrt{9x^2-12x+4}\)
i) \(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}+\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}\)
a) \(\left(\dfrac{1}{2-\sqrt{3}}-\dfrac{3}{\sqrt{7}-2}\right):\dfrac{2}{\sqrt{7}+\sqrt{3}}\)
b) \(\left(\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-1\right):\left(\sqrt{x}-x\right)+\dfrac{1}{x}\)
Với giá trị nào của x thì mỗi căn thức sau có nghĩa:
1 sqrt{9x^2-6x+1} 6 sqrt{x^2-16} 11 frac{1}{sqrt{9-12x+4x^2}}
2 sqrt{-x^2+2x-1} 7 sqrt{xleft(x+2right)} 12 sqrt{x^2-2x-3} sqrt{4x^2+3}
3frac{1}{sqrt{x+2sqrt{x-1}}} 8 sqrt{|x-1|-3} 13 sqrt{x^2-3}
4 sqrt{-|x+5|} 9 sqrt{|x|-1} sqrt{x^2+1}
5 sqrt{x^2-3} 10sqrt{x-2sqrt{x-1}}
Đọc tiếp
Với giá trị nào của x thì mỗi căn thức sau có nghĩa:
1 \(\sqrt{9x^2-6x+1}\) 6 \(\sqrt{x^2-16}\) 11 \(\frac{1}{\sqrt{9-12x+4x^2}}\)
2 \(\sqrt{-x^2+2x-1}\) 7 \(\sqrt{x\left(x+2\right)}\) 12 \(\sqrt{x^2-2x-3}\) \(\sqrt{4x^2+3}\)
3\(\frac{1}{\sqrt{x+2\sqrt{x-1}}}\) 8 \(\sqrt{|x-1|-3}\) 13 \(\sqrt{x^2-3}\)
4 \(\sqrt{-|x+5|}\) 9 \(\sqrt{|x|-1}\) \(\sqrt{x^2+1}\)
5 \(\sqrt{x^2-3}\) 10\(\sqrt{x-2\sqrt{x-1}}\)
câu1 ; gp trình
a, \(\sqrt{x-x}+x=3\)
b, \(\sqrt{x^2-4x+4}=x-2\)
câu 2 rút gọn
A = \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
B= \(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}\)
Giải các pt sau:
1, \(\sqrt{x^2+x+1}=2x+\sqrt{x^2-x+1}\)
2, \(2x^2+2x+6=2x\sqrt{x^2-x+1}+4\sqrt{3x+1}\)
3, \(\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\)
4, \(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2-2x+3}+\sqrt{x^2-x+2}\)
5, \(13\sqrt{x-1}+9\sqrt{x+1}=16x\)
Giải phương trình \(\sqrt{8x+1}+\sqrt{3x-5}=\sqrt{7x+4}+\sqrt{2x-2}\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
\(\sqrt{6-x}+\sqrt{x+2}=x^2-6x+13\)
\(\sqrt{x^2-x}+\sqrt{x^2+x-2}=0\)