-(\(\dfrac{3}{54}\)+\(\dfrac{3}{4}\))-(\(\dfrac{-3}{4}\)+\(\dfrac{2}{5}\))
Những câu hỏi liên quan
a)/x/-/\(\dfrac{-3}{4}\)/=/\(\dfrac{2}{5}\)/
b)/x+\(\dfrac{1}{3}\)/=/\(\dfrac{-2}{9}\)/
c)3\(^x\)*5+3\(^x\)=54
\(a,\Leftrightarrow\left|x\right|=\dfrac{2}{5}+\dfrac{3}{4}=\dfrac{23}{20}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=-\dfrac{23}{20}\end{matrix}\right.\\ b,\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{2}{9}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{9}\\x+\dfrac{1}{3}=-\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=-\dfrac{5}{9}\end{matrix}\right.\\ c,\Leftrightarrow3^x\cdot6=54\Leftrightarrow3^x=9=3^2\Leftrightarrow x=2\)
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1) \(\dfrac{x}{5}=\dfrac{y}{3}\) và x-y=20
2) \(\dfrac{x}{y}=\dfrac{3}{4}\) và x+y=90
3) \(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{3}\) và 2x+3y+4z=54
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU, TA ĐƯỢC :
`(x)/(3)=(y)/(4)=(x+y)/(3+4)=(90)/(7)`
`->` $\begin{cases}x=\dfrac{90}{7}.3=\dfrac{30}{7} \\ y=\dfrac{90}{7}.4=\dfrac{360}{7} \end{cases}$
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1)\(\dfrac{x}{5}=\dfrac{y}{3}\) áp dụng...ta đc:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{20}{2}=10\)
x=50
y=30
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ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU, TA ĐƯỢC :
`(x)/(5)=(y)/(3)=(x-y)/(5-3)=(20)/(2)=10`
`->` $\begin{cases} x=10.5=50\\ y=10.3=30\end{cases}$
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Xem thêm câu trả lời
dfrac{2sqrt{3}-3sqrt{2}}{sqrt{6}}-dfrac{2}{1-sqrt{3}}dfrac{4}{sqrt{6}+sqrt{2}}-dfrac{sqrt{54}+sqrt{2}}{sqrt{3}+1}dfrac{5+2sqrt{5}}{sqrt{5}}-dfrac{20}{5+sqrt{5}}-sqrt{20}Bài 2sqrt{25x^2-10x+1}sqrt{4x^2+8x+4}sqrt{x^2-3}+1xsqrt{7-2x}sqrt{x^2+7}sqrt{9x-27}+dfrac{1}{2}sqrt{4x-12}-9sqrt{dfrac{x-3}{9}}2
Đọc tiếp
\(\dfrac{2\sqrt{3}-3\sqrt{2}}{\sqrt{6}}-\dfrac{2}{1-\sqrt{3}}\)
\(\dfrac{4}{\sqrt{6}+\sqrt{2}}-\dfrac{\sqrt{54}+\sqrt{2}}{\sqrt{3}+1}\)
\(\dfrac{5+2\sqrt{5}}{\sqrt{5}}-\dfrac{20}{5+\sqrt{5}}-\sqrt{20}\)
Bài 2
\(\sqrt{25x^2-10x+1}=\sqrt{4x^2+8x+4}\)
\(\sqrt{x^2-3}+1=x\)
\(\sqrt{7-2x}=\sqrt{x^2+7}\)
\(\sqrt{9x-27}+\dfrac{1}{2}\sqrt{4x-12}-9\sqrt{\dfrac{x-3}{9}}=2\)
\(2,\\ a,PT\Leftrightarrow\sqrt{\left(5x-1\right)^2}=\sqrt{4\left(x+1\right)^2}\\ \Leftrightarrow\left|5x-1\right|=2\left|x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=2\left(x+1\right)\\1-5x=2\left(x+1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3\\7x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{7}\end{matrix}\right.\)
\(b,ĐK:x^2-3\ge0\\ PT\Leftrightarrow\sqrt{x^2-3}=x-1\\ \Leftrightarrow x^2-3=x^2-2x+1\\ \Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ c,ĐK:x\le\dfrac{7}{2}\\ PT\Leftrightarrow7-2x=x^2+7\\ \Leftrightarrow x^2+2x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge3\\ PT\Leftrightarrow3\sqrt{x-3}+\dfrac{1}{2}\cdot2\sqrt{x-3}-9\cdot\dfrac{1}{3}\sqrt{x-3}=2\\ \Leftrightarrow\sqrt{x-3}=2\\ \Leftrightarrow x-3=4\Leftrightarrow x=7\left(tm\right)\)
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Bài 1:
d: Ta có: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}-\dfrac{20}{5+\sqrt{5}}-\sqrt{20}\)
\(=\sqrt{5}+2-5+\sqrt{5}-2\sqrt{5}\)
=-3
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tim xdfrac{5}{6}x-dfrac{3}{4}dfrac{-1}{4}+dfrac{2}{3}65x−434−1+32-1dfrac{1}{2}-dfrac{2}{3}xdfrac{5}{6}-left(dfrac{-2}{5}right)−121−32x65−(5−2)left(dfrac{4}{5}:x+1,5right):dfrac{2}{3}-1,5(54:x+1,5):32−1,5dfrac{4}{3}x-dfrac{2}{3}dfrac{1}{4}-x34x−3241−xgiup minh nhe minh dang can gap
Đọc tiếp
tim x
\dfrac{5}{6}x-\dfrac{3}{4}=\dfrac{-1}{4}+\dfrac{2}{3}65x−43=4−1+32
-1\dfrac{1}{2}-\dfrac{2}{3}x=\dfrac{5}{6}-\left(\dfrac{-2}{5}\right)−121−32x=65−(5−2)
\left(\dfrac{4}{5}:x+1,5\right):\dfrac{2}{3}=-1,5(54:x+1,5):32=−1,5
\dfrac{4}{3}x-\dfrac{2}{3}=\dfrac{1}{4}-x34x−32=41−x
giup minh nhe minh dang can gap
\(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)
Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}-\dfrac{x-60}{56}-\dfrac{x-60}{55}-\dfrac{x-60}{54}=0\)
\(\Leftrightarrow x-60=0\)
hay x=60
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Tính
a) \(2\sqrt[3]{24}-5\sqrt[3]{81}+4\sqrt[3]{192}\)
b) \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
c) \(\dfrac{\sqrt[3]{4}+\sqrt[3]{2}}{3}-\dfrac{1}{\sqrt[3]{2}+1}\)
b: Ta có: \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}\cdot\sqrt[3]{4}\)
\(=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54\cdot4}\)
=3-6
=-3
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tính1.sqrt{147}+sqrt{54}-4sqrt{27}2.sqrt{28}-4sqrt{63}+7sqrt{112}3.sqrt{49}-5sqrt{28}+dfrac{1}{2}sqrt{63}4.left(2sqrt{6}-4sqrt{3}-dfrac{1}{4}sqrt{8}right).3sqrt{6}5.(2sqrt{1dfrac{9}{16}}-5sqrt{5dfrac{1}{16}}):sqrt{16}6.left(sqrt{48}-3sqrt{27}-sqrt{147}right):sqrt{3}7.left(sqrt{50}-3sqrt{49}right):sqrt{2}-sqrt{162}:sqrt{2}8.left(2sqrt{1dfrac{9}{10}}-sqrt{5dfrac{1}{10}}right):sqrt{10}9.2sqrt{dfrac{16}{3}}-3sqrt{dfrac{1}{27}}-6sqrt{dfrac{4}{75}}10.2sqrt{27}-6sqrt{dfrac{4}{3}}+dfrac{3}{5}sqrt{75}11....
Đọc tiếp
tính
1.\(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
2.\(\sqrt{28}-4\sqrt{63}+7\sqrt{112}\)
3.\(\sqrt{49}-5\sqrt{28}+\dfrac{1}{2}\sqrt{63}\)
4.\(\left(2\sqrt{6}-4\sqrt{3}-\dfrac{1}{4}\sqrt{8}\right).3\sqrt{6}\)
5.(\(2\sqrt{1\dfrac{9}{16}}-5\sqrt{5\dfrac{1}{16}}\)):\(\sqrt{16}\)
6.\(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt{3}\)
7.\(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt{2}-\sqrt{162}:\sqrt{2}\)
8.\(\left(2\sqrt{1\dfrac{9}{10}}-\sqrt{5\dfrac{1}{10}}\right):\sqrt{10}\)
9.\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
10.\(2\sqrt{27}-6\sqrt{\dfrac{4}{3}}+\dfrac{3}{5}\sqrt{75}\)
11.\(\dfrac{\sqrt{18}}{\sqrt{2}}-\dfrac{\sqrt{12}}{\sqrt{3}}\)
12.\(\dfrac{\sqrt{27}}{\sqrt{3}}+\dfrac{\sqrt{98}}{\sqrt{2}}-\sqrt{175}:\sqrt{7}\)
13.\(\left(\dfrac{\sqrt{8}}{\sqrt{2}}-\dfrac{\sqrt{180}}{\sqrt{5}}\right).\sqrt{5}-\sqrt{\dfrac{81}{11}}.\sqrt{11}\)
14.\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
15.\(\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
16.\(\left(1+\sqrt{5}-\sqrt{3}\right)\left(1+\sqrt{5}+\sqrt{3}\right)\)
Giải phương trình:
a) \(\sqrt{x}+\sqrt{2-x}=\dfrac{3x^2-2x+3}{x^2+1}\)
b) \(x^3-11x^2+36x-18=4\sqrt[4]{27x-54}\)
c) \(16x^4+5=6\sqrt[3]{4x^3+x}\)
d) \(\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}=\dfrac{2}{x}\)
b, \(đk:x\ge2\)
Xét x=2 thay vào pt thấy không thỏa mãn => x>2 hay 27x-54>0
\(x^3-11x+36x-18=4\sqrt[4]{27x-54}\)
\(\Leftrightarrow27x^3-297x^2+972x-486=4\sqrt[4]{\left(27x-54\right).81.81.81}\le189+27x\) (cosi với 4 số dương, dấu = xảy ra khi x=5)
\(\Leftrightarrow x^3-11x^2+35x-25\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)^2\le0\) (*)
Có \(\left\{{}\begin{matrix}x>2\\\left(x-5\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-5\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left(x-1\right)\left(x-5\right)^2\ge0\) (2*)
Từ (*) và (2*) ,dấu = xra khi x=5 (thỏa mãn)
Vây pt có nghiệm duy nhất x=5
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c,Có \(6\sqrt[3]{4x^3+x}=16x^4+5>0\)
\(\Leftrightarrow4x^3+x>0\)
Có: \(16x^4+5=6\sqrt[3]{4x^3+x}\le2\left(4x^3+x+2\right)\) (theo cosi với 3 số dương,dấu = xảy ra khi \(x=\dfrac{1}{2}\))
\(\Leftrightarrow16x^4-8x^3-2x+1\le0\)
\(\Leftrightarrow\left(2x-1\right)^2\left(4x^2+2x+1\right)\le0\) (*)
(tương tự câu b) Dấu = xảy ra khi \(x=\dfrac{1}{2}\)(thỏa mãn)
Vậy....
d) Đk: \(x\ge\dfrac{3}{4}\)
Áp dụng bđt cosi:
\(\sqrt{2x-1}\le\dfrac{2x-1+1}{2}=x\)
\(\Rightarrow\dfrac{1}{\sqrt{2x-1}}\ge\dfrac{1}{x}\) (*)
\(\sqrt[4]{4x-3}\le\dfrac{4x-3+1+1+1}{4}=x\)
\(\dfrac{\Rightarrow1}{\sqrt[4]{4x-3}}\ge\dfrac{1}{x}\) (2*)
Từ (*) và (2*) \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}\ge\dfrac{2}{x}\)
Dấu = xảy ra khi x=1 (tm)
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`a)\sqrtx+\sqrt{2-x}=(3x^2-2x+3)/(x^2+1)`
`đk:0<=x<=2`
`pt<=>sqrtx-1+\sqrt{2-x}-1=(3x^2-2x+3)/(x^2+1)-2`
`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x^2-2x+1)/(x^2+1)`
`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x-1)^2/(x^2+1)`
`<=>(x-1)((x-1)/(x^2+1)+1/(sqrt{2-x}+1)-1/(sqrtx+1))=0`
`<=>x-1=0<=>x=1`
Vậy `S={1}`
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Tính:
a) \(\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{11}{2}\right);\)
b) \(\dfrac{28}{15}\cdot\dfrac{1}{4^2}.3+\left(\dfrac{8}{15}-\dfrac{69}{60}\cdot\dfrac{5}{23}\right):\dfrac{51}{54}.\)
\(a.\)
\(\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{11}{2}\right)\)
\(=\dfrac{17}{8}:\left(\dfrac{27+44}{8}\right)=\dfrac{17}{8}:\dfrac{71}{8}=\dfrac{17}{8}\cdot\dfrac{8}{71}=\dfrac{17}{71}\)
\(b.\)
\(\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{69}{60}\cdot\dfrac{5}{23}\right):\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{1}{4}\right):\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8\cdot4-15}{60}\right):\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\dfrac{17}{60}:\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{16}\cdot3+\dfrac{17}{60}\cdot\dfrac{54}{51}\)
\(=\dfrac{7}{20}+\dfrac{3}{10}\)
\(=\dfrac{7+3\cdot2}{20}=\dfrac{13}{20}\)
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