b: \(=x-4\sqrt{x}+3\sqrt{x}-12=\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)\)

Fpai

#)Thắc mắc ?

Bạn ơi ! chỗ kia là \(\sqrt{x}-7hay\sqrt{x+7}\)thế ???????????????

#)Giải :

\(5\sqrt{x-1}-\sqrt{x-7}=3x-4\)

ĐKXĐ : \(x\ge1\)

Đặt \(\hept{\begin{cases}\sqrt{x-1}=a\ge0\\\sqrt{x+7=b>0}\end{cases}\Rightarrow3x-4}=\frac{25a^2-b^2}{8}\)

Phương trình trở thành : 

\(5a-b=\frac{25a^2-b^2}{8}\Leftrightarrow\left(5a-b\right)\left(5a+b\right)=8\left(5a-b\right)\)

 \(\Leftrightarrow\orbr{\begin{cases}5a-b=0\\5a+b=8\end{cases}\Leftrightarrow\orbr{\begin{cases}5\sqrt{x-1}=\sqrt{x+7}\\5\sqrt{x-1}+\sqrt{x+7}=8\end{cases}}}\)

\(TH1:5\sqrt{x+1}=\sqrt{x+7}\Leftrightarrow25\left(x-1\right)=x+7\Rightarrow x=\frac{4}{3}\)

\(TH2:5\sqrt{x-1}+\sqrt{x+7}=8\)

\(\Leftrightarrow5\sqrt{x-1}-5+\sqrt{x+7}-3=0\)

\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{x-1}+1}+\frac{x-2}{\sqrt{x-7}+3}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{x-1}+1}+\frac{1}{\sqrt{x-7}+3}\right)=0\)

\(\Rightarrow x=2\)

a: ΔKAC vuông tại K

=>\(\hat{KAC}+\hat{KCA}=90^0\)

=>\(\hat{KAC}=90^0-30^0=60^0\)

Xét ΔKAC vuông tại K có sin C=\(\frac{AK}{AC}\)

=>AC=3:sin30=6(cm)

ΔAKC vuông tại K

=>\(KA^2+KC^2=AC^2\)

=>\(KC^2=6^2-3^2=36-9=27\)

=>\(KC=3\sqrt3\) (cm)

b: Xét ΔAKB vuông tại K có cot B=\(\frac{BK}{AK}\)

=>\(BK=AK\cdot\cot B\)

Xét ΔAKC vuông tại K có \(\cot C=\frac{CK}{AK}\)

=>\(CK=AK\cdot\cot C\)

BK+CK=BC

=>\(AK\cdot\left(\cot B+\cot C\right)=BC\)

=>\(AK=\frac{BC}{\cot B+\cot C}\)

a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x-\sqrt{x}-1}{x-2\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{x-5}{x-\sqrt{x}-2}\right)\)

\(=\dfrac{x-x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{1}\)

\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\)

Bài 2:ĐKXĐ: a>=0; a<>4

a: \(\frac{\sqrt{a}-2}{\sqrt{a}+2}+\frac{\sqrt{a}+2}{\sqrt{a}-2}-\frac{4a}{4-a}\)

\(=\frac{\left(\sqrt{a}-2\right)^2+\left(\sqrt{a}+2\right)^2+4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{a-4\sqrt{a}+4+a+4\sqrt{a}+4+4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{6a+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{2\left(3a+4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

Ta có: \(M=\left(\frac{\sqrt{a}-2}{\sqrt{a}+2}+\frac{\sqrt{a}+2}{\sqrt{a}-2}-\frac{4a}{4-a}\right):\frac{3a+4}{\sqrt{a}+2}\)

\(=\frac{2\left(3a+4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{3a+4}{\sqrt{a}+2}\)

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

b: M<-1

=>M+1<0

=>\(\frac{2+\sqrt{a}-2}{\sqrt{a}-2}<0\)

=>\(\frac{\sqrt{a}}{\sqrt{a}-2}<0\)

=>\(\sqrt{a}-2<0\)

=>\(\sqrt{a}<2\)

=>0<=a<4

c: Để M nguyên thì 2⋮\(\sqrt{a}-2\)

=>\(\sqrt{a}-2\in\left\lbrace1;-1;2;-2\right\rbrace\)

=>\(\sqrt{a}\in\left\lbrace3;1;4;0\right\rbrace\)

=>a∈{0;1;9;16}

Bài 1:

a: \(A=\frac{x\cdot\sqrt{2x}+1}{x-1}-\frac{x+\sqrt{2x}}{x-1}\)

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

\(B=\sqrt2\cdot\sqrt{2+\sqrt3}-\frac{2}{\sqrt3+1}\)

\(=\sqrt{4+2\sqrt3}-\frac{2\left(\sqrt3-1\right)}{\left(\sqrt3+1\right)\left(\sqrt3-1\right)}\)

\(=\sqrt3+1-\left(\sqrt3-1\right)=2\)

b: Khi x=2 thì \(A=\sqrt{2\cdot2}-1=2-1=1\)

c: A=B

=>\(\sqrt{2x}-1=2\)

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

=>2x=9

=>x=9/2(nhận)

Ta có: \(\hat{MDN}=\hat{MPN}=\hat{MBN}=90^0\)

=>M,D,N,P,B cùng thuộc đường tròn đường kính MN

Em cần câu nào nhỉ?

em cần câu c và d ạ :<

\(a,ĐK:x\in R\\ PT\Leftrightarrow\sqrt{\left(x-2\right)^2}=3\Leftrightarrow\left|x-2\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)