(1,5 điểm) Tìm $x$ biết
a) $\dfrac{3}{4}-\left(x-\dfrac{2}{3}\right)=1 \dfrac{1}{3}$;
b) $\Big| \dfrac{1}{3}-x \Big|=\dfrac{2}{5}$;
c) $\dfrac{x-1}{3}=\dfrac{27}{x-1}$.
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Tìm $x$, biết :
a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
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(1,5 điểm) a) Chứng minh đẳng thức: $\left( 2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1} \right).\left( 2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1} \right)=1.$
b) Rút gọn biểu thức $A=\left( \dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2} \right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}$ với $x>0;$ $x\ne 4$.
a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
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Ta có đẳng thức : (2−3+√3√3+1).(2+3−√3√3−1)=1
xét vế trái ta có :(2−3+√3√3+1).(2+3−√3√3−1) =
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a) ta co \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right).\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)=1\)
b) ta co \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)
\(A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)^2}\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)
\(A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
Vay \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
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29 tháng 1 2022 lúc 11:20
Tìm x biết:a,3dfrac{1}{2}-dfrac{1}{2}xdfrac{2}{3}b,dfrac{1}{3}+dfrac{2}{3}:x-7c,dfrac{1}{3}x+dfrac{2}{5}left(x-1right)0d,left(2x-3right)left(6-2xright)0e,x:dfrac{3}{4}+dfrac{1}{4}-dfrac{2}{3}f,dfrac{-2}{3}-dfrac{1}{3}left(2x-5right)dfrac{3}{2}g,2left|dfrac{1}{2}x-dfrac{1}{3}right|-dfrac{3}{2}dfrac{1}{4}h,dfrac{3}{4}-2.left|2x-dfrac{2}{3}right|2i,left(-0,6x-dfrac{1}{2}right).dfrac{3}{4}-left(-1right)dfrac{1}{3}j,left(3x-1right)left(-dfrac{1}{2}x+5right)0k,dfrac{1}{4}+dfrac{1}{3}:left(2x-1right)-5...
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Tìm x biết:
\(a,3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\)
\(b,\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(c,\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(d,\left(2x-3\right)\left(6-2x\right)=0\)
\(e,x:\dfrac{3}{4}+\dfrac{1}{4}=-\dfrac{2}{3}\)
\(f,\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(g,2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(h,\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(i,\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(j,\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(k,\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(l,\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(m,3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(n,60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(p,-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(q,3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\)
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
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Tìm x,y biết :
a) \(\left|3.x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}.y+\dfrac{3}{5}\right|\)= 0
b)\(\left|\dfrac{3}{2}.x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}.y-\dfrac{1}{2}\right|\le0\)
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
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28 tháng 2 2022 lúc 15:04
Tìm x biết:
\(a,\left(x-\dfrac{3}{4}\right)+50\%=\dfrac{1}{6}\)
\(b,\dfrac{1}{2}x-\dfrac{5}{6}x=\dfrac{7}{2}\)
\(c,\left(4-x\right)\left(3x+5\right)=0\)
\(d,\dfrac{x}{16}=\dfrac{50}{32}\)
\(e,\left(2x-3\right)+\dfrac{3}{2}=-\dfrac{1}{4}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
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Tìm x, biết:
a) \(\dfrac{2}{3}\)x - \(\dfrac{1}{2}\)x = \(\left(-\dfrac{7}{12}\right)\) . \(1\dfrac{2}{5}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2\) = \(\dfrac{9}{4}\)
c) (1,25 - \(\dfrac{4}{5}\)x)3 = -125
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
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a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
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(1,5 điểm)Thực hiện phép tính:
a) $\dfrac{3}{4}+1 \dfrac{2}{3}-\dfrac{7}{9}$;
b) $\dfrac{1}{7} \cdot \left(\dfrac{4}{3}\right)^2-\dfrac{1}{7}: \dfrac{9}{11}$;
c) $0,5 \cdot \dfrac{4}{9}+\left(\dfrac{1}{3}-1,5\right)^2-\left(\dfrac{2}{3}\right)^9:\left(\dfrac{2}{3}\right)^7$.
Bài 1.(2,5 điểm)Tìm x, biết:a) left(2dfrac{1}{3}+3dfrac{1}{2}right).x-4dfrac{1}{6}+3dfrac{1}{2}b) left(1dfrac{1}{3}+3dfrac{1}{2}right).x4dfrac{1}{6}-3dfrac{1}{2}c) dfrac{1}{3}-dfrac{7}{8}.xdfrac{1}{4}d) dfrac{3}{2}.x+dfrac{1}{7}dfrac{7}{8}.dfrac{64}{49}e) 5dfrac{1}{2}-left(dfrac{1}{4}.x+dfrac{2}{5}right)25%
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Bài 1.(2,5 điểm)Tìm x, biết:
a) \(\left(2\dfrac{1}{3}+3\dfrac{1}{2}\right).x=-4\dfrac{1}{6}+3\dfrac{1}{2}\)
b) \(\left(1\dfrac{1}{3}+3\dfrac{1}{2}\right).x=4\dfrac{1}{6}-3\dfrac{1}{2}\)
c) \(\dfrac{1}{3}-\dfrac{7}{8}.x=\dfrac{1}{4}\)
d) \(\dfrac{3}{2}.x+\dfrac{1}{7}=\dfrac{7}{8}.\dfrac{64}{49}\)
e) \(5\dfrac{1}{2}-\left(\dfrac{1}{4}.x+\dfrac{2}{5}\right)=25\%\)
c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)
\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)
d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)
hay \(x=\dfrac{2}{3}\)
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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
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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