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a) | x-1 | + | x^2+3 | = 0
b) | x-1 | + | x^2 -1 | = 0
c) | x^2 -4 | + | x | = 2
mn ưi giúp mk vs ạ :((
a) Ta có: \(\left|x-1\right|+\left|x^2+3\right|=0\)
\(\Leftrightarrow\left|x-1\right|=-\left|x^2+3\right|\)
Mà \(\hept{\begin{cases}\left|x-1\right|\ge0\\-\left|x^2+3\right|\le0\end{cases}\left(\forall x\right)}\)
Dấu "=" xảy ra khi: \(\left|x-1\right|=-\left|x^2+3\right|=0\)
\(\Rightarrow x^2=-3\) => vô lý
Vậy PT vô nghiệm
b) Ta có: \(\left|x-1\right|+\left|x^2-1\right|=0\)
\(\Leftrightarrow\left|x-1\right|=-\left|x^2-1\right|\)
Mà \(\hept{\begin{cases}\left|x-1\right|\ge0\\-\left|x^2-1\right|\le0\end{cases}\left(\forall x\right)}\)
Dấu "=" xảy ra khi: \(\left|x-1\right|=-\left|x^2-1\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\x^2=1\end{cases}}\Rightarrow x=1\)
Vậy x = 1
\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\left(đk:x\ge1\right)\)
\(< =>\sqrt{x-2\sqrt{x-1}}^2=\left(\sqrt{x-1}-1\right)^2\)
\(< =>x-2\sqrt{x-1}=x-1+1-2\sqrt{x-1}\)
\(< =>x-2\sqrt{x-1}+2\sqrt{x-1}=x< =>x=x\)
Vậy phương trình trên thỏa mãn với mọi \(x\ge1\)
ĐKXĐ : \(x\ge1\)
Bình phương 2 vế lên ta có :
\(x-2\sqrt{x-1}=\left(\sqrt{x-1}-1\right)^2\)
\(\Leftrightarrow x-2\sqrt{x-1}=x-1-2\sqrt{x-1}+1\)
\(\Leftrightarrow x-2\sqrt{x-1}=x-2\sqrt{x-1}\)
\(\Leftrightarrow0x=0\)( luôn đúng với mọi \(x\ge1\))
Vậy ...............
a: Ta có: AD=DE=EC
mà AD+DE+EC=3a
nên \(AD=DE=EC=a\)
mà AB=a
nên AB=AD=DE=EC=a và DC=2a
Áp dụng định lí Pytago vào ΔABD vuông tại A, ta được:
\(BD^2=BA^2+AD^2\)
\(\Leftrightarrow BD^2=a^2+a^2=2a^2\)
hay \(BD=a\sqrt{2}\)
Ta có: \(\dfrac{DE}{DB}=\dfrac{a}{a\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
mà \(\dfrac{DB}{DC}=\dfrac{a\sqrt{2}}{2a}=\dfrac{\sqrt{2}}{2}\)
nên \(\dfrac{DE}{DB}=\dfrac{DB}{DC}\)
b: Xét ΔBDE và ΔCDB có
\(\dfrac{DE}{DB}=\dfrac{DB}{DC}\)
\(\widehat{BDC}\) chung
Do đó: ΔBDE\(\sim\)ΔCDB
a: Ta có: \(M=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
b: Ta có: \(x=\dfrac{8}{\sqrt{5}-1}-\dfrac{8}{\sqrt{5}+1}\)
\(=2\left(\sqrt{5}+1\right)-2\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}+2-2\sqrt{5}+2\)
=4
Thay x=4 vào M, ta được:
\(M=\dfrac{2}{4+2+1}=\dfrac{2}{7}\)
Ta có \(P=\sum\dfrac{1}{\sqrt{2a^2+5ab+2b^2}}\le\sum\dfrac{1}{\sqrt{9ab}}=\dfrac{1}{3}\sum\dfrac{1}{\sqrt{ab}}\le\dfrac{1}{6}\sum\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{2}{3}\).
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\dfrac{3}{2}\)
\(3x^2-2\left(m-3\right)x-2m+3=0\)
\(\Delta'=\left(m-3\right)^2-\left(-2m+3\right)=m^2-4m+6=\left(m-2\right)^2+2>0\)
\(\Rightarrow\) PT luôn có 2 nghiệm phân biệt
Áp dụng hệ thức Vi-et ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-3\right)\\x_1x_2=-2m+3\end{matrix}\right.\)
Khi đó ta có:
\(P=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\\ =4\left(m-3\right)^2-2\left(-2m+3\right)=4m^2-20m+30\)
\(=\left(4m^2-20m+25\right)+5=\left(2m-5\right)^2+5\ge5\)
Dấu = xảy ra \(\Leftrightarrow2m-5=0\Leftrightarrow m=\dfrac{5}{2}\)
a) Ta có: \(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-\left(a-\sqrt{a}+2\sqrt{a}-2\right)}{\sqrt{a}}\)
\(=2+\dfrac{3a+3\sqrt{a}-a+\sqrt{a}-2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\left(a+2\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
b) Ta có: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2\left(a-\sqrt{a}+1\right)}{\sqrt{a}}>0\forall a\) thỏa mãn ĐKXĐ
hay P>6
Bài 86:
a: Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để Q>0 thì \(\sqrt{a}-2>0\)
hay a>4
6, \(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=\sqrt{x}\left(\sqrt{y}+2\right)-3\left(\sqrt{y}+2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{y}+2\right)\)
7, \(7+2\sqrt{10}=7+2\sqrt{5.2}=5+2\sqrt{5.2}+2=\left(\sqrt{5}+\sqrt{2}\right)^2\)
8, \(5-2\sqrt{6}=5-2\sqrt{2.3}=3-2\sqrt{2.3}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\)
9, \(\sqrt{x^2-y^2}-x+y=\sqrt{\left(x-y\right)\left(x+y\right)}-\left(x-y\right)\)
\(=\sqrt{x-y}\left(\sqrt{x+y}-\sqrt{x-y}\right)\)
10, \(3x-2\sqrt{x}=\sqrt{x}\left(3\sqrt{x}-2\right)\)
1, \(\sqrt{xy}-x=\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)\)
2, \(x+y-2\sqrt{xy}=\left(\sqrt{x}-\sqrt{y}\right)^2\)
3, \(x\sqrt{y}-y\sqrt{x}=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)
4, \(2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}=2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)\)
\(=\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)\)
5, \(\sqrt{35}-\sqrt{14}=\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)\)