1. Tìm x
a) (x+1)2 = 3
b) (x-1)x+2 = (x-1)x+6
c) ( x+\(\frac{1}{2}\))2 = \(\frac{4}{25}\)
2. Tìm x
a) \(\sqrt{x}\)= 4
b) \(\sqrt{x+1}\)= 5
c) 5(x+2)(x-3) = 1
d) (2x - 1)12 = (x+1)12
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
MH
25 tháng 9 2023
a) \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-1\right)}=-\left(x+3+x-1-6\right)\)\(\left(Đk:x\ge1\right)\)
\(\left(\sqrt{x-1}+\sqrt{x+3}\right)^2+\sqrt{x-1}+\sqrt{x-3}-6=0\)
\(\left(\sqrt{x-1}+\sqrt{x+3}+3\right)\left(\sqrt{x-1}+\sqrt{x+3}-2\right)=0\)
Đến đây em xét các trường hợp rồi bình phương lên là được nha
b) \(\sqrt{3x-2}+\sqrt{x-1}=3x-2+x-1-6+2\sqrt{\left(3x-2\right)\left(x-1\right)}\left(Đk:x\ge1\right)\)
\(\left(\sqrt{3x-2}+\sqrt{x-1}\right)^2-\left(\sqrt{3x-2}+\sqrt{x-1}\right)-6=0\)
\(\left(\sqrt{3x-2}+\sqrt{x-1}-3\right)\left(\sqrt{3x-2}+\sqrt{x-1}+2\right)=0\)
Đến đây em xét các trường hợp rồi bình phương lên là được nha
AH
Akai Haruma
Giáo viên
25 tháng 9 2023
a/ ĐKXĐ: $x\geq 1$
Đặt $\sqrt{x-1}=a; \sqrt{x+3}=b$ thì pt trở thành:
$a+b+2ab=6-(a^2+b^2)$
$\Leftrightarrow a^2+b^2+2ab+a+b-6=0$
$\Leftrightarrow (a+b)^2+(a+b)-6=0$
$\Leftrightarrow (a+b-2)(a+b+3)=0$
Hiển nhiên do $a\geq 0; b\geq 0$ nên $a+b+3>0$. Do đó $a+b-2=0$
$\Leftrightarrow a+b=2$
Mà $b^2-a^2=(x+3)-(x-1)=4$
$\Leftrightarrow (b-a)(b+a)=4\Leftrightarrow (b-a).2=4\Leftrightarrow b-a=2$
$\Rightarrow \sqrt{x+3}=b=(a+b+b-a):2=(2+2):2=2$
$\Leftrightarrow x=1$ (tm)
19 tháng 7 2018
\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)
\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)
\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)
\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)
19 tháng 7 2018
\(4,A=x+\sqrt{x}+1\)
\(A=\left(\sqrt{x}\right)^2+2.\frac{1}{2}.\sqrt{x}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(A=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)
Dấu "=" xảy ra khi :
\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)
Vậy Min A = 3/4 khi căn x = -1/2
6 tháng 4 2023
a: =>x/27+1=-2/3
=>x/27=-5/3
=>x=-45
b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)
=>x=221/50
c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)
=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)
=>x=-37/24
e: =>-3/7x=84/45
=>x=-196/45
f: =>11/10x=-2/3
=>x=-20/33
25 tháng 8 2021
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
26 tháng 8 2021
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
4 tháng 12 2019
1.
ĐK: \(-1\le x\le4\)
Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)
\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)
\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)
2.
ĐK:\(x\ge4\)
Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)
\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)
\(PT\Leftrightarrow t=2x-12+t^2-2x\)
\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2019
a/ ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{2\left(x-1\right)}{x}+3\)
Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)
\(\frac{2}{a}-a=2a^2+3\Leftrightarrow2a^3+a^2+3a-2=0\)
\(\Leftrightarrow\left(2a-1\right)\left(a^2+a+2\right)=0\Leftrightarrow a=\frac{1}{2}\)
\(\Rightarrow\sqrt{\frac{x-1}{x}}=\frac{1}{2}\Leftrightarrow4\left(x-1\right)=x\)
b/ ĐKXĐ: ...
\(\Leftrightarrow3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{3\left(x-1\right)}{2x}+10\)
Đặt \(\sqrt{\frac{x-1}{2x}}=a>0\)
\(\frac{3}{a}+4a=3a^2+10\Leftrightarrow3a^3-4a^2+10a-3=0\)
\(\Leftrightarrow\left(3a-1\right)\left(a^2-a+3\right)=0\Leftrightarrow a=\frac{1}{3}\)
\(\Leftrightarrow\sqrt{\frac{x-1}{2x}}=\frac{1}{3}\Leftrightarrow9\left(x-1\right)=2x\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2019
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{\frac{x}{3-2x}}+5\sqrt{\frac{3-2x}{x}}=\frac{4\left(3-2x\right)}{x}+5\)
Đặt \(\sqrt{\frac{3-2x}{x}}=a>0\)
\(\frac{1}{a}+5a=4a^2+5\Leftrightarrow4a^3-5a^2+5a-1=0\)
\(\Leftrightarrow\left(4a-1\right)\left(a^2-a+1\right)=0\Leftrightarrow a=\frac{1}{4}\)
\(\Leftrightarrow\sqrt{\frac{3-2x}{x}}=\frac{1}{4}\Leftrightarrow16\left(3-2x\right)=x\)
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)
\(a^2-2a=3\Leftrightarrow a^2-2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=3\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\frac{x-1}{x}}=3\Leftrightarrow x-1=9x\)
1, \(a,\left(x+1\right)^2=3\)
\(\Rightarrow x+1=\pm\sqrt{3}\)
\(\Rightarrow x=\pm\sqrt{3}-1\)
\(b,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Rightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^4-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^4-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x-1=\pm1\Rightarrow x=2or\text{ }x=0\end{cases}}\)
\(c,\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow x+\frac{1}{2}=\pm\sqrt{\frac{4}{25}}\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}}\)
2, \(a,\sqrt{x}=4\)
\(\Rightarrow\sqrt{x}=\sqrt{16}\)
\(\Rightarrow x=16\)
\(b,\sqrt{x+1}=5\)
\(\Rightarrow\sqrt{x+1}=\sqrt{25}\)
\(\Rightarrow x+1=25\)
\(\Rightarrow x=24\)
\(\Rightarrow5^{\left(x+2\right)\left(x+3\right)}=1\)
\(\Rightarrow5^{\left(x+2\right)\left(x+3\right)}=5^0\)
\(\Rightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)
\(d,\left(2x-1\right)^{12}=\left(x+1\right)^{12}\)
\(\Rightarrow\left(2x-1\right)^{12}\div\left(x+1\right)^{12}=1\)
\(\Rightarrow\)