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28 tháng 8 2023
1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)
=>4x+8=3x-1
=>x=-9
2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)
=>8x-4=5x-7
=>3x=-3
=>x=-1
3: ĐKXD: x>=0
\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=>\(x+\sqrt{x}-6=x-1\)
=>căn x=-1+6=5
=>x=25
4: ĐKXĐ: x>=0
PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
=>x-2*căn x-3=x-4
=>-2căn x-3=-4
=>2căn x+3=4
=>2căn x=1
=>căn x=1/2
=>x=1/4
T
10 tháng 8 2021
Làm a, c là tiêu biểu thôi, bài b đơn giản.
a) \(\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=\sqrt{x-1}-1\)
ĐKXĐ: $x\ge 1.$ Do $VT\ge 0 \Rightarrow VT\ge 0 \to x\ge 2.$
Ta có \(VT=\sqrt{\left[\sqrt{x-1}-1\right]^2}=\left|\sqrt{x-1}-1\right|=VP\) (vì \(\sqrt{x-1}-1=VP\ge0.\))
Vậy phương trình có vô số nghiệm.
c) Ta có:
\(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=2\)
ĐKXĐ: $x\ge 1.$
Ta có: \(VT=\sqrt{\left(\sqrt{x-1}+1\right)^2}=\left|\sqrt{x-1}+1\right|=\sqrt{x-1}+1.\)
(vì $\sqrt{x-1}+1>0\forall x\ge 1.$)
Ta có: \(\sqrt{x-1}+1=2\Rightarrow x=2.\) (thỏa mãn)
10 tháng 8 2021
b: Ta có: \(\sqrt{36x^2-12x+1}=5\)
\(\Leftrightarrow\left|6x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=6\\6x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
15 tháng 1 2022
a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)
b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)
=>x<1
12 tháng 1 2022
\(a,P=\dfrac{-x+2\sqrt{x}-1+x-2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}:\dfrac{2\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ P=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\\ \Rightarrow P=\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{5-\sqrt{5}}{5}\\ c,\dfrac{P}{\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{1}{\sqrt{x}-1}\le\dfrac{1}{0-1}=-1\)
Vậy \(\left(\dfrac{P}{\sqrt{x}}\right)_{max}=-1\Leftrightarrow x=0\)
AH
Akai Haruma
Giáo viên
7 tháng 7 2021
1.
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$A=|x+2|+|x+3|=|x+2|+|-x-3|\geq |x+2-x-3|=1$
Vậy GTNN của $A$ là $1$. Giá trị này đạt tại $(x+2)(-x-3)\geq 0$
$\Leftrightarrow (x+2)(x+3)\leq 0$
$\Leftrightarrow -3\leq x\leq -2$
AH
Akai Haruma
Giáo viên
7 tháng 7 2021
2. ĐKXĐ: $x\geq 1$
\(B=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\sqrt{(x-1)+2\sqrt{x-1}+1}+\sqrt{(x-1)-2\sqrt{x-1}+1}\)
\(=\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|\)
\(=|\sqrt{x-1}+1|+|1-\sqrt{x-1}|\geq |\sqrt{x-1}+1+1-\sqrt{x-1}|=2\)
Vậy gtnn của $B$ là $2$. Giá trị này đạt tại $(\sqrt{x-1}+1)(1-\sqrt{x-1})\geq 0$
$\Leftrightarrow 1-\sqrt{x-1}\geq 0$
$\Leftrightarrow 0\leq x\leq 2$
24 tháng 10 2021
1) ĐKXĐ: \(x\ge-2\)
\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)
\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)
\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)
2) ĐKXĐ: \(x\ge0\)
\(pt\Leftrightarrow2x+3=x^2\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
3) \(pt\Leftrightarrow\sqrt{\left(5x+2\right)^2}=1\Leftrightarrow\left|5x+2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=1\\5x+2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)
\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)
\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)
5) ĐKXĐ: \(x\ge2\)
\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)
\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)
Vậy \(S=\varnothing\)
8 tháng 2 2021
1) Ta có: \(P=\left(\dfrac{\sqrt{x}}{2\sqrt{x}-2}+\dfrac{3-\sqrt{x}}{2x-2}\right):\left(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}+2}{x\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\sqrt{x}}{2\left(\sqrt{x}-1\right)}+\dfrac{3-\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3-\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x-2\sqrt{x}+1+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x+\sqrt{x}+3-\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+3}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-\sqrt{x}+3}\)
\(=\dfrac{\left(x+3\right)\left(x+\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+3\right)}\)
KL
10 tháng 5 2021
B=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{6}{\sqrt{x}-1}-\dfrac{\sqrt{x}+15}{x+2\sqrt{x}-3}\) Bạn ơi giúp mk câu này vs ạ !
GH
3 tháng 7 2023
Với \(x\ge0;x\ne4\) có:
\(A=\dfrac{x+2}{x-2\sqrt{x}+\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\\ =\dfrac{x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
a
\(P=A:B=\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{\left(4\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}\)
b
\(P^2=P+2\\ \Leftrightarrow P^2-P-2=0\\ \Leftrightarrow P^2-2P+P-2=0\\ \Leftrightarrow P\left(P-2\right)+\left(P-2\right)=0\\ \Leftrightarrow\left(P-2\right)\left(P+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}P=2\\P=-1\end{matrix}\right.\)
Với P = 2 có:
\(\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}=2\\ \Leftrightarrow2\left(\sqrt{x}+1\right)=4\sqrt{x}+1\\ \Leftrightarrow2\sqrt{x}+2-4\sqrt{x}-1=0\\\Leftrightarrow -2\sqrt{x}+1=0\\\Leftrightarrow-2\sqrt{x}=-1\\\Leftrightarrow \sqrt{x}=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{1}{4} \)
Với P = -1 có:
\(\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}=-1\\ \Leftrightarrow-\sqrt{x}-1-4\sqrt{x}-1=0\\ \Leftrightarrow-5\sqrt{x}=2\\ \Leftrightarrow\sqrt{x}=-\dfrac{2}{5}\left(loại\right)\)
Vậy để \(P^2=P+2\) thì \(x=\dfrac{1}{4}\)
3 tháng 7 2023
a: P=A:B
\(=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}\)
b: P^2=P+2
=>P^2-P-2=0
=>(P-2)(P+1)=0
=>P=2(nhận) hoặc P=-1(loại)
=>\(4\sqrt{x}+1=2\sqrt{x}+2\)
=>2căn x=1
=>x=1/4
ĐKXĐ: x>=1
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|=2\)
Ta có \(\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|=2\)
Dấu "=" xảy ra khi \(\left(\sqrt{x-1}+1\right)\left(1-\sqrt{x-1}\right)\ge0\)
<=> x=<2. Kết hợp với ĐKXĐ => 1=<x=<2