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Sửa đề: x-4
\(A=\dfrac{x-2\sqrt{x}+x+4\sqrt{x}+4+2x+8}{x-4}=\dfrac{4x+2\sqrt{x}+12}{x-4}\)
a, Ta có :
\(P=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{2x+\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}-2}\)sử dụng tam thức bậc 2 khai triển biểu thức trên tử nhé
\(=\frac{\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)
\(Q=\frac{\left(\sqrt{x}\right)^3-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}=x-1\)
b, Ta có : \(P=Q\)hay \(2\sqrt{x}+1=x-1\Leftrightarrow-x+2\sqrt{x}+2=0\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-3=0\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
TH1 : \(\sqrt{x}=1+\sqrt{3}\Leftrightarrow x=\left(1+\sqrt{3}\right)^2=1+2\sqrt{3}+3=4+2\sqrt{3}\)
TH2 : \(\sqrt{x}=1-\sqrt{3}\Leftrightarrow x=\left(1-\sqrt{3}\right)^2=1-2\sqrt{3}+3=4-2\sqrt{3}\)
Vậy \(x=4+2\sqrt{3};x=4-2\sqrt{3}\)thì P = Q
んuﾘ ｲ giải pt vô tỉ không xét ĐK là tai hại :))
\(P=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{2x-4\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)
\(Q=\frac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{\left(x\sqrt{x}-\sqrt{x}\right)+\left(2x-2\right)}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}=\frac{\left(x-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=x-1\)
Để P = Q thì \(2\sqrt{x}+1=x-1\)( x ≥ 1 ; x ≠ 4 )
<=> \(x-2\sqrt{x}-2=0\)
<=> \(\left(\sqrt{x}-1\right)^2-3=0\)
<=> \(\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
<=> \(\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=4+2\sqrt{3}\left(tm\right)\\x=4-2\sqrt{3}\left(ktm\right)\end{cases}}\)
Vậy với \(x=4+2\sqrt{3}\)thì P = Q
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
a: \(B=\dfrac{2x+3\sqrt{x}+9-x+3\sqrt{x}}{x-9}=\dfrac{x+9}{x-9}\)
b: \P=A:B
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\cdot\dfrac{x-9}{x+9}=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{x+9}>=\dfrac{-1\cdot3}{9}=\dfrac{-1}{3}\)
Dấu = xảy ra khi x=0
Câu 1:
a) Khi x =16 (t.m ĐKXĐ) thì B có giá trị là:
\(B=\dfrac{16-6\cdot4}{4-1}=\dfrac{-8}{3}\)
b) Ta có:
\(A=\dfrac{25\sqrt{x}+6}{x-36}-\dfrac{\sqrt{x}-1}{6-\sqrt{x}}+\dfrac{2\sqrt{x}}{\sqrt{x}+6}=\dfrac{25\sqrt{x}+6}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-6\right)}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\dfrac{25\sqrt{x}+6+x+5\sqrt{x}-6+2x-12\sqrt{x}}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\dfrac{3x+18\sqrt{x}}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-6}\)
c) Ta có:
\(T=\sqrt{A\cdot B}=\sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}-6}\cdot\dfrac{x-6\sqrt{x}}{\sqrt{x}-1}}=\sqrt{\dfrac{3x\left(\sqrt{x}-6\right)}{\left(\sqrt{x}-6\right)\left(\sqrt{x}-1\right)}}=\sqrt{\dfrac{3\left(x-1\right)+3}{\sqrt{x}-1}}=\sqrt{3\left(\sqrt{x}+1\right)+\dfrac{3}{\sqrt{x}-1}}=\sqrt{3\left(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\right)+6}\overset{Cosi}{\ge}\sqrt{3\cdot2+6}=2\sqrt{3}\)
Dấu = xảy ra \(\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(t.m\right)\)
Gọi vận tốc dự định của hai bố con bạn Dũng là x(km/h)(x>0).Đổi: 10 phút =\(\dfrac{1}{6}\)(h)
thời gian dự định đi về quê là \(\dfrac{60}{x}\)(h)
vận tốc đi trên \(\dfrac{1}{3}\)quãng đường là đường xấu hai bố con bạn Dũng là \(x-10\)(km/h)
Thời gian thực tế đi về quê là \(\dfrac{\dfrac{1}{3}\cdot60}{x-10}+\dfrac{\dfrac{2}{3}\cdot60}{x}\)(h)
Vì hai bố con bạn Dũng đã về tới quê chậm mất 10 phút so với dự kiến
Nên ta có pt sau:
\(\left(\dfrac{\dfrac{1}{3}\cdot60}{x-10}+\dfrac{\dfrac{2}{3}\cdot60}{x}\right)-\dfrac{1}{6}=\dfrac{60}{x}\)
⇔\(\dfrac{20}{x-10}+\dfrac{40}{x}-\dfrac{1}{6}=\dfrac{60}{x}\)
⇔\(20x+40\left(x-10\right)-\dfrac{1}{6}x\left(x-10\right)=60\left(x-10\right)\)
⇔\(-\dfrac{1}{6}x^2+\dfrac{5}{3}x+200=0\)
⇒\(\left[{}\begin{matrix}x=40\left(n\right)\\x=-30\left(l\right)\end{matrix}\right.\)
Vậy ......
a, Với \(x>0;x\ne4;x\ne9\)
\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)
b, Ta có : A = -2 hay
\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)
\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)
\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)
bình phương 2 vế ta có :
\(x=\left(2x+3\right)^2=4x^2+12x+9\)
\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
\(A=\sqrt{\dfrac{18-3\sqrt{3}}{11}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
\(=\dfrac{2\sqrt{11\left(18-3\sqrt{3}\right)}-11\sqrt{6}-11\sqrt{2}}{22}\)
b: \(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a: \(=-4+2\sqrt{5}-\sqrt{5}+2+\sqrt{5}=2\sqrt{5}-2\)
b: \(B=\dfrac{2\sqrt{x}+4+6\sqrt{x}-3-2\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}}{6\sqrt{x}+4}\)
\(=\dfrac{\left(6\sqrt{x}+1\right)\cdot\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(6\sqrt{x}+4\right)}\)
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