Cho biểu thức P=\(\dfrac{3x+\sqrt{x}}{x+\sqrt{x}}+\dfrac{3\left(x-\sqrt{x}+1\right)}{x\sqrt{x}+1}\)
a) rút gọn P
b)chứng minh P<4
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a: ĐKXĐ: x=0; x<>1
\(M=\left(2+\sqrt{x}\right)\left(1-2\sqrt{x}-x+1+\sqrt{x}+x\right)\)
\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)
b: Sửa đề: P=1/M
P=1/4-x=-1/x-4
Để P nguyên thì x-4 thuộc {1;-1}
=>x thuộc {5;3}
có `cos α=1/2`
`=>cos^2 α=1/4`
Mà `cos^2 α +sin^2 α=1`
`=>1/4+sin^2 α=1`
`=>sin^2 α=1-1/4=3/4`
\(=>sin\alpha=\dfrac{\sqrt{3}}{2}\) (vì `sin α` >0)
ta có `sin α : cos α=tan α`
\(=>tan\alpha=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)
ta có `tan α * cot α =1`
\(=>\sqrt{3}\cdot cot\alpha=1\\ =>cot\alpha=\dfrac{1}{\sqrt{3}}\)
tương tự ta có
\(\left\{{}\begin{matrix}sin\beta=\dfrac{\sqrt{2}}{2}\\cos\beta=1\\cot\beta=1\end{matrix}\right.\)
sin a=12/13
cos^2a=1-(12/13)^2=25/169
=>cosa=5/13
tan a=12/13:5/13=12/5
cot a=1:12/5=5/12
sin b=căn 3/2
cos^2b=1-(căn 3/2)^2=1/4
=>cos b=1/2
tan b=căn 3/2:1/2=căn 3
cot b=1/căn 3
\(A=\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+y}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}{\left(x-y\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{x+\sqrt{xy}+y}{x+y}\cdot\dfrac{x+\sqrt{xy}-\sqrt{xy}+y}{x-y}\)
\(=\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{x+\sqrt{xy}+y}{x-y}\)
\(=\dfrac{\sqrt{xy}+y-x-\sqrt{xy}-y}{x-y}=\dfrac{-x}{x-y}\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y}\right)-2x\) (với \(x\ne y,x,y\ge0\))
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}+\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{\sqrt{y}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}+\sqrt{y}-\sqrt{x}}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{2\sqrt{y}}{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)-2x\)
\(P=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\cdot2\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)}{\sqrt{x}-\sqrt{y}}-2x\)
\(P=\dfrac{2\sqrt{x}\left(y-x\right)-2x\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-2x\sqrt{x}-2x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(P=\dfrac{2y\sqrt{x}-4x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
6:
AC^2=AH*AB
=>AH(AH+16)=15^2=225
=>AH^2+16AH-225=0
=>(AH+25)(AH-9)=0
=>AH=9cm
=>BA=16+9=25cm
BC=căn 25^2-15^2=20cm
S BCA=1/2*CA*CB=1/2*15*20=150cm2
a) \(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\)
\(=\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\)
\(=2\sqrt{5}:\sqrt{6}\)
\(=\sqrt{\dfrac{20}{6}}\)
\(=\sqrt{\dfrac{10}{3}}\)
\(=\dfrac{\sqrt{30}}{3}\)
b) \(\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)
\(=\left(\sqrt{\dfrac{1}{7}}-4\sqrt{\dfrac{1}{7}}+\sqrt{7}\right):\sqrt{7}\)
\(=\left(-3\sqrt{\dfrac{1}{7}}+\sqrt{7}\right):\sqrt{7}\)
\(=4\sqrt{\dfrac{1}{7}}:\sqrt{7}\)
\(=\dfrac{4}{7}\)
c) \(\left(\sqrt{325}-\sqrt{117}+2\sqrt{208}\right):\sqrt{13}\)
\(=\left(5\sqrt{13}-3\sqrt{13}+8\sqrt{13}\right):\sqrt{13}\)
\(=10\sqrt{13}:\sqrt{13}\)
\(=10\)
d) \(\left(\dfrac{1}{3}\sqrt{\dfrac{1}{2}}-\dfrac{2}{3}\sqrt{\dfrac{3}{2}}+\dfrac{2}{7}\sqrt{\dfrac{1}{6}}\right):\left(\dfrac{2}{7}\sqrt{\dfrac{1}{8}}\right)\)
\(=\left(\dfrac{\sqrt{2}}{6}-\dfrac{\sqrt{6}}{3}+\dfrac{\sqrt{6}}{21}\right):\dfrac{\sqrt{2}}{14}\)
\(=\dfrac{-12\sqrt{6}+7\sqrt{2}}{42}:\dfrac{\sqrt{2}}{14}\)
\(=\dfrac{7-12\sqrt{3}}{3}\)
7:
a: \(A=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1\)
\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)
b: A=2
=>a-căn a-2=0
=>(căn a-2)(căn a+1)=0
=>căn a-2=0
=>a=4
c; A=a-căn a+1/4-1/4=(căn a-1/2)^2-1/4>=-1/4
Dấu = xảy ra khi a=1/4
ĐKXĐ: x>=0; x<>1
PT =>\(\dfrac{\left(\sqrt{x}+3\right)\left(-2x+6\right)}{\left(\sqrt{x}-1\right)^2}=0\)
=>6-2x=0
=>x=3
a: \(P=\dfrac{3\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{3}{\sqrt{x}+1}=\dfrac{3\sqrt{x}+4}{\sqrt{x}+1}\)
b: \(P-4=\dfrac{3\sqrt{x}+4-4\sqrt{x}-4}{\sqrt{x}+1}=-\dfrac{\sqrt{x}}{\sqrt{x}+1}< 0\)
=>P<4