\(\sqrt{7-\frac{1}{5}x}\) 2) \(\sqrt{\frac{4}{3} 2x}\) \(\sqrt{\frac{2}{x^2 \frac{1}{2}}}\) 4) \(\sqrt{\frac{-2}{x^2 \frac{1}{3}}}\) \(\sqrt{\frac{-5}{x^2 4x}}\) 6) \(\sqrt{\frac{5}{x^2 4x}}\) \(\s...

giải pt a) \(3\sqrt{x}+\frac{3}{2\sqrt{x}}=2x+\frac{1}{2x}-7\) b) \(5\sqrt{x}+\frac{5}{2\sqrt{x}}=2x+\frac{1}{2x}+4\) c) \(\sqrt{2x^2+8x+5}+\sqrt{2x^2-4x+5}=6\sqrt{x}\) d) \(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\) e) \(x^2+2x\sqrt{x-\frac{1}{x}}=3x+1\) f) \(x^2-6x+x\sqrt{\frac{x^2-6}{x}}-6=0\) g) \(\frac{3x^2}{3+\sqrt{x}}+6+2\sqrt{x}=5x\) h)...

0

JE

Julian Edward

26 tháng 10 2019

#Toán lớp 10

5

NV

Nguyễn Việt Lâm

Giáo viên

27 tháng 10 2019

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

\(\Rightarrow x+\frac{1}{4x}=a^2-1\)

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

NV

Nguyễn Việt Lâm

Giáo viên

27 tháng 10 2019

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

BÀI 1: RÚT GỌN 1)\(\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-1}\) 2)\(\sqrt{7+2\sqrt{10}}+2\sqrt{\frac{1}{5}}-\frac{1}{\sqrt{5}-2}\) 3)\(\frac{3}{\sqrt{3}-1}+\sqrt{\frac{4}{3}}-\sqrt{8+2\sqrt{5}}\) 4)\(3\sqrt{\frac{16x}{81}}+\frac{5}{4}\sqrt{\frac{4x}{25}}-\frac{2}{x}\sqrt{\frac{9a^3}{4}}\) 5)\(\frac{1}{3}\sqrt{3a}-\frac{2}{3}\sqrt{\frac{27a}{4}}+\frac{5}{a}\sqrt{\frac{12a^3}{5}}\) BÀI 2: GIẢI PHƯƠNG...

PM

Phương Minh

3 tháng 9 2019

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,...

0

N

Nam

22 tháng 9 2020

Bài 1: Tìm điều kiện để các phân thức sau có nghĩa a)\(\frac{x-1}{x+1}b)\frac{2x+1}{-3x+5}c)\frac{3x-1}{x^2-4}d)\frac{x-1}{x^2+4}e)\frac{x-1}{\left(x-2\right)\left(x+3\right)}g)\frac{x-1}{x+2}:\frac{x}{x+1}\) Bài 2 :Tìm điều kiện để các căn thức sau có...

0

PM

Phương Minh

31 tháng 7 2019

2

NT

nguyễn thị mai linh

30 tháng 3 2020

https://i.imgur.com/iX7y3qX.jpg

NT

nguyễn thị mai linh

30 tháng 3 2020

https://i.imgur.com/GMDpx0f.jpg

a. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}\) f. \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7-3}}{2-\sqrt{x+3}}\) b. \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}\) g. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\) c. \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}\) h. \(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}\) d. \(\lim\limits_{x\rightarrow1}\frac{3x-2\sqrt{4x^2-x-2}}{x^2-3x+2}\) k....

TP

Trần Phương Thảo

15 tháng 3 2020

#Toán lớp 11

1

MC

meo con

15 tháng 3 2020

a) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}=\lim\limits_{x\rightarrow0}\frac{2x}{2x\left(\sqrt{1+2x}+1\right)}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{1+2x}+1}=\frac{1}{2}\)

b) \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\frac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}[4\left(\sqrt{9+x}+3\right)=24\)

c) \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\frac{1}{\sqrt{x+7}+3}=\frac{1}{6}\)

d) \(\lim\limits_{x\rightarrow1}\frac{3x-2-\sqrt{4x^2-x-2}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\frac{\left(3x-2\right)^2-\left(4x^2-4x-2\right)}{(x^2-3x+2)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(5x-6\right)}{\left(x-1\right)\left(x-2\right)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\frac{1}{2}\\ \\\\ \\ \\ \\ \)

e)\(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\frac{2x+7-\left(x^2-8x+16\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x-9\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{x-9}{\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=-8\)

f) \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}=\lim\limits_{x\rightarrow1}\frac{(2x-2)\left(2+\sqrt{x+3}\right)}{\left(1-x\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(2+\sqrt{x+3}\right)}{\sqrt{2x+7}+3}=\frac{-4}{3}\)

g) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}=\lim\limits_{x\rightarrow0}\frac{x^2\left(\sqrt{x^2+16}+4\right)}{x^2\left(\sqrt{x^2+1}+1\right)}=4\)

h)

\(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-3}{x-4}+\lim\limits_{x\rightarrow4}\frac{3-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{1}{\sqrt{x+5}+4}+\lim\limits_{x\rightarrow4}\frac{8-2x}{\left(x-4\right)\left(3+\sqrt{2x+1}\right)}=\frac{1}{7}-\frac{1}{3}=\frac{-4}{21}\)

k) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)

VT

Võ Thiên Long

30 tháng 7 2019

Giải các phương trinh sau

a. \(\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\) b.\(\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)

c\(\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}=4}\)

d. \(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)

e. \(\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)

4

TA

Thuý An Nguyễn Thị

30 tháng 7 2019

Đề câu c ptrinh = 4 là phải riêng ra chứ

VT

Vy Thị Hoàng Lan ( Toán Học )

30 tháng 7 2019

\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)

\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)

\(\Rightarrow3x+2=2\left(x+2\right)\)

\(\Rightarrow3x+2=2x+4\)

\(\Rightarrow3x-2x=4-2\)

\(\Rightarrow x=2\)

\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)

\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)

\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)

\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)

\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)

\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)

\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)

\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)

\(\Rightarrow2\sqrt{x-2}=4\)

\(\Rightarrow\sqrt{x-2}=2\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)

\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)

\(\Rightarrow x+4=1-2x+2\sqrt{\left(1-2x\right)\left(1-x\right)}+1-x\)

\(\Rightarrow x+4=2-3x+2\sqrt{1-3x+2x^2}\)

\(\Rightarrow x+4-2+3x=2\sqrt{1-3x+2x^2}\)

\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)

\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)

\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)

\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)

\(\Rightarrow2x^2+7x=0\)

\(\Rightarrow x\left(2x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)

\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)

\(\frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{2x\left(\sqrt{3}-1\right)}{3-1}=\sqrt{5}+1\)

\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)

\(\Rightarrow\sqrt{5}x+\sqrt{3}x-\sqrt{3x}+x=\sqrt{5}+1\)

\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)

\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)

\(\Rightarrow x=1\)

TB

trần bảo nhi

17 tháng 10 2020

1) \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)

2) \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)

3) \(\sqrt{x^2-6x+9}=2x\)

4) \(\sqrt{4x^2+1}=2x-1\)

5) \(\sqrt{x^2-4x+4}=\sqrt{x^2-2x+1}\)

1

NL

Nguyễn Lê Phước Thịnh

17 tháng 10 2020

1) Ta có: \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\cdot\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)

\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{25}{4}\cdot2}+12\right)\)

\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{50}{4}}+12\right)\)

\(=-12\sqrt{2}+12-\frac{5\sqrt{2}}{2}-12\)

\(=\frac{-24\sqrt{2}-5\sqrt{2}}{2}\)

\(=\frac{-29\sqrt{2}}{2}\)

2) Ta có: \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)

\(=\frac{26\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}+\frac{4}{2-\sqrt{3}}\)

\(=\frac{26\left(5-2\sqrt{3}\right)}{25-12}+\frac{4\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=2\left(5-2\sqrt{3}\right)+4\left(2+\sqrt{3}\right)\)

\(=10-4\sqrt{3}+8+4\sqrt{3}\)

\(=18\)

3) ĐK để phương trình có nghiệm là: x≥0

Ta có: \(\sqrt{x^2-6x+9}=2x\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x\)

\(\Leftrightarrow\left|x-3\right|=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\\x-3=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3-2x=0\\x-3+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x-3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=3\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Vậy: S={1}

4) ĐK để phương trình có nghiệm là: \(x\ge\frac{1}{2}\)

Ta có: \(\sqrt{4x^2+1}=2x-1\)

\(\Leftrightarrow\left(\sqrt{4x^2+1}\right)^2=\left(2x-1\right)^2\)

\(\Leftrightarrow4x^2+1=4x^2-4x+1\)

\(\Leftrightarrow4x^2+1-4x^2+4x-1=0\)

\(\Leftrightarrow4x=0\)

hay x=0(loại)

Vậy: S=∅