Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
H24
22 tháng 9 2020 lúc 22:50
1, frac{x}{2}-frac{3-x}{3}frac{2x+2}{5}
2,1-frac{3-x}{3}frac{2x+2}{5}-frac{2-x}{4}
3,frac{2}{3}x+1x-5
4, 2x-x2 0
5,frac{4x}{x+1}+frac{x+3}{x}6
6, frac{x-1}{x-3}+frac{2x+2}{x-2}8
7, sqrt{x-1}sqrt{2}
8, sqrt{2x-1}sqrt{x}-4
Đọc tiếp
1, \(\frac{x}{2}-\frac{3-x}{3}=\frac{2x+2}{5}\)
2,1-\(\frac{3-x}{3}=\frac{2x+2}{5}-\frac{2-x}{4}\)
3,\(\frac{2}{3}x+1=x-5\)
4, 2x-x2 =0
5,\(\frac{4x}{x+1}+\frac{x+3}{x}=6\)
6, \(\frac{x-1}{x-3}+\frac{2x+2}{x-2}=8\)
7, \(\sqrt{x-1}=\sqrt{2}\)
8, \(\sqrt{2x-1}=\sqrt{x}-4\)
Giải phương trình
1.\(\sqrt{2x-3}-\sqrt{5-2x}=3x^2-12x+14\)
2.\(x^2+2x+15=6\sqrt{4x+5}\)
3.\(x^2-5x-8=2\sqrt{x-2}\)
4.\(\sqrt{x+1+\sqrt{x+\frac{3}{4}}}=x+1\)
\(\sqrt{12-\frac{12}{x^2} }+\sqrt{x^2+\frac{12}{x} }=x^2+\frac{25}{2} \)
\(\frac{\sqrt[3]{10-x}+\sqrt[3]{8-x}}{\sqrt[3]{10-x}-\sqrt[3]{8-x}}=9-x \)
\(4x^2+\sqrt{2x+1}+5=12x\)
a)sqrt{x^2+2x+10}+x^2+2x+80
b)15x-2x^2-5sqrt{2x^2-15x+11}
c)sqrt{9x^2+45}+sqrt{16x^2+80}+3sqrt{frac{x^2+5}{16}}-frac{1}{4}sqrt{frac{25x^2+15}{9}}9
d)3x^2+21x+18+2sqrt{x^2+7x+7}2
e)sqrt{x^2+3x+2}-2sqrt{2x^2+6x+2}-sqrt{2}
f)sqrt{x-1}+sqrt{x+3}-sqrt{x^2+2x-3}-10
Đọc tiếp
a)\(\sqrt{x^2+2x+10}+x^2+2x+8=0\)
b)\(15x-2x^2-5=\sqrt{2x^2-15x+11}\)
c)\(\sqrt{9x^2+45}+\sqrt{16x^2+80}+3\sqrt{\frac{x^2+5}{16}}-\frac{1}{4}\sqrt{\frac{25x^2+15}{9}}=9\)
d)\(3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)
e)\(\sqrt{x^2+3x+2}-2\sqrt{2x^2+6x+2}=-\sqrt{2}\)
f)\(\sqrt{x-1}+\sqrt{x+3}-\sqrt{x^2+2x-3}-1=0\)
1) Giải phương trình:
a)\(\sqrt{3x+1}=2x+4\)
b)\(\sqrt{\left(x-3\right)^2}=2x^2+x-3\)
c)\(\frac{3x+\sqrt{5}}{\sqrt{2}}-\sqrt{8}=0\)
d)\(\frac{x^2-\sqrt{8}}{\sqrt{2}}+\sqrt{18}=0\)
e)\(2\sqrt{x}=3\sqrt{x}-2\)
Giúp mk vs mk đang cần gấp
Bài 1: Tính
a) sqrt{9-sqrt{17}}cdotsqrt{9+sqrt{17}}
b) sqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}
c) left(sqrt{3-sqrt{5}}+sqrt{3+sqrt{5}}right)^2
d) sqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}cdotsqrt{2-sqrt{2+sqrt{2+sqrt{3}}}}
e) sqrt{frac{8+sqrt{15}}{2}}+sqrt{frac{8-sqrt{15}}{2}}
Bài 2: Giải pt:
a) sqrt{x+3-4sqrt{x-1}}+sqrt{x+8-6sqrt{x-1}}1
b) sqrt{x+sqrt{x-11}}+sqrt{x-sqrt{x-11}}4
c) sqrt{x+2+3sqrt{2x-5}}+sqrt{x-2-sqrt{2x-5}}2sqrt{2}
d) sqrt{x-4}+sqrt{6-x}x^2-10x+27
e) sqrt{2x+1}+sqrt{...
Đọc tiếp
Bài 1: Tính
a) \(\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)
b) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
c) \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
d) \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
e) \(\sqrt{\frac{8+\sqrt{15}}{2}}+\sqrt{\frac{8-\sqrt{15}}{2}}\)
Bài 2: Giải pt:
a) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
b) \(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)
c) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
d) \(\sqrt{x-4}+\sqrt{6-x}=x^2-10x+27\)
e) \(\sqrt{2x+1}+\sqrt{17-2x}=x^4-8x^3+17x^2-8x+22\)
f) \(\sqrt{x+x^2}+\sqrt{x-x^2}=x+1\)
g) \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=5\)
Bài 3: Cho biểu thức:
P= \(\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}}\)
a) Rút gon P
b) Tìm x để P đạt GTNN, tìm GTNN đó.
c) Tìm x \(\in\) Z để P \(\in\) Z
@Nguyễn Văn Đạt@Akai Haruma Help me please~~~~ Giải thích cẩn thân hộ với.
Giải phương trình:
a) 2sqrt{x^2-4}-36sqrt{x-2}-sqrt{x+2}
b) frac{sqrt{x-2016}-1}{x-2016}+frac{sqrt{y-2017}-1}{y-2017}+frac{sqrt{z-2018}-1}{z-2018}frac{3}{4}
c) sqrt{3+sqrt{3+x}}x
d) sqrt{6x^2+1}sqrt{2x-3}+x^2
e) sqrt{x^2+3x+5}+sqrt{x^2-2x+5}5sqrt{x}
f) sqrt{x^2+3x}+2sqrt{x+2}2x+sqrt{x+frac{6}{x}+5}
Đọc tiếp
Giải phương trình:
a) \(2\sqrt{x^2-4}-3=6\sqrt{x-2}-\sqrt{x+2}\)
b) \(\frac{\sqrt{x-2016}-1}{x-2016}+\frac{\sqrt{y-2017}-1}{y-2017}+\frac{\sqrt{z-2018}-1}{z-2018}=\frac{3}{4}\)
c) \(\sqrt{3+\sqrt{3+x}}=x\)
d) \(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
e) \(\sqrt{x^2+3x+5}+\sqrt{x^2-2x+5}=5\sqrt{x}\)
f) \(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\frac{6}{x}+5}\)
\(\frac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}-\frac{\sqrt{x}-4}{\sqrt{x}+1}-\frac{\sqrt{x}+8}{\sqrt{x}-4}\)
Tìm ĐKXĐ:
a; \(\sqrt[4]{\frac{2}{-7+3x}}\)
b; \(\sqrt{x-1}+\frac{\sqrt[3]{x+1}}{\sqrt{5-x}}\)
c; \(\sqrt[8]{2x-1}-\sqrt[3]{3-5x}\)
d; \(\sqrt{\frac{3x-6-2x}{\sqrt[3]{1-x}}}\)