Cho biểu thức $A=\Big(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\Big) \, : \, \Big(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\Big)$ với $x>0; \, x\ne 4; \, x\ne 9$.
a) Rút gọn biểu thức $A$.
b) Tìm $x$ để $A = - 2$.
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1619 và 825
1619 = (24)19 = 276
825 = (23)25 = 275
Vì 275 < 276 nên
1619 > 825
a: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-\left(a-4\right)}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: P>1/6
=>P-1/6>0
=>\(\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}>0\)
=>\(\dfrac{6\left(\sqrt{a}-2\right)-3\sqrt{a}}{18\sqrt{a}}>0\)
=>\(6\left(\sqrt{a}-2\right)-3\sqrt{a}>0\)
=>\(3\sqrt{a}-12>0\)
=>\(\sqrt{a}>4\)
=>a>16
1: Thay x=9 vào A, ta được:
\(A=\dfrac{3\cdot3}{3+2}=\dfrac{9}{5}\)
2: \(B=\dfrac{x+4}{x-4}-\dfrac{2}{\sqrt{x}-2}\)
\(=\dfrac{x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}\)
\(=\dfrac{x+4-2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
3: \(A-B< \dfrac{3}{2}\)
=>\(\dfrac{3\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}< \dfrac{3}{2}\)
=>\(\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{3}{2}< 0\)
=>\(\dfrac{4\sqrt{x}-3\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+2\right)}< 0\)
=>\(\dfrac{\sqrt[]{x}-6}{2\left(\sqrt{x}+2\right)}< 0\)
=>\(\sqrt{x}-6< 0\)
=>\(\sqrt{x}< 6\)
=>0<=x<36
mà x là số nguyên dương lớn nhất thỏa mãn
nên x=35
\(P=\left(1+\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-1+\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}=\dfrac{2\left(\sqrt[]{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{2}{\sqrt{x}}\)
a: \(\left(\sqrt{\dfrac{4}{3}}+\sqrt{3}\right)\cdot\sqrt{6}\)
\(=\sqrt{\dfrac{4}{3}\cdot6}+\sqrt{3\cdot6}\)
\(=\sqrt{8}+\sqrt{18}=2\sqrt{2}+3\sqrt{2}=5\sqrt{2}\)
b: \(\left(1-2\sqrt{5}\right)^2=\left(2\sqrt{5}-1\right)^2\)
\(=\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot1+1\)
\(=21-4\sqrt{5}\)
c: \(2\sqrt{3}-\sqrt{27}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
d: \(\sqrt{45}-\sqrt{20}+\sqrt{5}\)
\(=3\sqrt{5}-2\sqrt{5}+\sqrt{5}\)
\(=4\sqrt{5}-2\sqrt{5}=2\sqrt{5}\)
Đây là toán nâng cao chuyên đề bội ước, cấu trúc thi chuyên, thi học sinh giỏi các cấp. Hôm nay, Olm sẽ hướng dẫn các em giải chi tiết dạng này như sau.
45 = 32.51
Vì 45 là ước chung lớn nhất của A = 3a.53 và 33.5b
Nên \(\left\{{}\begin{matrix}3^a=3^2\\5^b=5^1\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
Vậy a + b = 2 + 1 = 3
b: \(\left(x-\dfrac{1}{3}\right)^3=-\dfrac{8}{27}\)
=>\(\left(x-\dfrac{1}{3}\right)^3=\left(-\dfrac{2}{3}\right)^3\)
=>\(x-\dfrac{1}{3}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}+\dfrac{1}{3}=-\dfrac{1}{3}\)
c: \(\left(5x+1\right)^2=\dfrac{36}{49}\)
=>\(\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{6}{7}-1=-\dfrac{1}{7}\\5x=-\dfrac{6}{7}-1=-\dfrac{13}{7}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{7}:5=-\dfrac{1}{35}\\x=-\dfrac{13}{7}:5=-\dfrac{13}{35}\end{matrix}\right.\)
d: \(\left(\dfrac{1}{3}-\dfrac{3}{2}x\right)^2=2\dfrac{1}{4}\)
=>\(\left(\dfrac{3}{2}x-\dfrac{1}{3}\right)^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}\dfrac{3}{2}x-\dfrac{1}{3}=\dfrac{3}{2}\\\dfrac{3}{2}x-\dfrac{1}{3}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=\dfrac{3}{2}+\dfrac{1}{3}=\dfrac{11}{6}\\\dfrac{3}{2}x=-\dfrac{3}{2}+\dfrac{1}{3}=-\dfrac{7}{6}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{11}{6}:\dfrac{3}{2}=\dfrac{11}{6}\cdot\dfrac{2}{3}=\dfrac{11}{9}\\x=-\dfrac{7}{6}:\dfrac{3}{2}=-\dfrac{7}{6}\cdot\dfrac{2}{3}=-\dfrac{7}{9}\end{matrix}\right.\)
e: \(\left(\dfrac{4}{5}\right)^{2x+5}=\dfrac{256}{625}\)
=>\(\left(\dfrac{4}{5}\right)^{2x+5}=\left(\dfrac{4}{5}\right)^4\)
=>2x+5=4
=>2x=4-5=-1
=>\(x=-\dfrac{1}{2}\)
g: \(\left(\dfrac{1}{3}\right)^{x+1}+\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{1}{12}\)
=>\(\left(\dfrac{1}{3}\right)^x\cdot\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^x\cdot\dfrac{1}{9}=\dfrac{1}{12}\)
=>\(\left(\dfrac{1}{3}\right)^x\left(\dfrac{1}{3}+\dfrac{1}{9}\right)=\dfrac{1}{12}\)
=>\(\left(\dfrac{1}{3}\right)^x=\dfrac{1}{12}:\dfrac{4}{9}=\dfrac{1}{12}\cdot\dfrac{9}{4}=\dfrac{3}{4\cdot4}=\dfrac{3}{16}\)
=>\(x=log_{\dfrac{1}{3}}\left(\dfrac{3}{16}\right)\)
Scorpio is a water sign and is known for its intense and passionate nature. People born under this sign are often mysterious, powerful, and determined. They value loyalty and are not afraid to stand up for themselves. However, they can also be jealous and possessive.
Của mik là cung Bạch Dương nha:
Aries is a zodiac sign of the Fire element along with Leo and Sagittarius. They have enthusiastic, creative personalities and are not afraid of challenges. Aries is also one of the four Ruling signs along with Libra, Capricom and Cancer. They have leadership, decisiveness ang steadfastness
a: \(A=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt[]{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\sqrt{x}+3}=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{4x}{\sqrt{x}-3}\)
b: A=-2
=>\(4x=-2\left(\sqrt{x}-3\right)=-2\sqrt{x}+6\)
=>\(4x+2\sqrt{x}-6=0\)
=>\(2x+\sqrt{x}-3=0\)
=>\(\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)
mà \(2\sqrt{x}+3>=3>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-1=0\)
=>x=1(nhận)