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\(a)3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right):2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
\(=\dfrac{7}{2}.\dfrac{4}{49}-\dfrac{88}{27}:\left(\dfrac{-42}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{-220}{567}\)
\(=\dfrac{382}{567}\)
các phần con lại dễ nên bn tự lm đi nhé mk bn lắm
Chúc bạn học tốt!
a/ \(x+\dfrac{3}{5}=\dfrac{4}{7}\)
\(x=\dfrac{4}{7}-\dfrac{3}{5}\)
\(x=-\dfrac{1}{35}\)
Vậy ....
b/ \(x-\dfrac{5}{6}=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+\dfrac{5}{6}\)
\(x=1\)
Vậy ....
c/\(-\dfrac{5}{7}-x=\dfrac{-9}{10}\)
\(x=\dfrac{-5}{7}-\dfrac{-9}{10}\)
\(x=\dfrac{13}{70}\)
Vậy .....
d/ \(\dfrac{5}{7}-x=10\)
\(x=\dfrac{5}{7}-10\)
\(x=\dfrac{-65}{7}\)
Vậy ...
e/ \(x:\left(\dfrac{1}{9}-\dfrac{2}{5}\right)=\dfrac{-1}{2}\)
\(x:\dfrac{-13}{45}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}.\dfrac{-13}{45}\)
\(x=\dfrac{13}{90}\)
Vậy ....
f/ \(\left(\dfrac{-3}{5}+1,25\right)x=\dfrac{1}{3}\)
\(0,65.x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:0,65\)
\(x=\dfrac{20}{39}\)
Vậy ....
g/ \(\dfrac{1}{3}x+\left(\dfrac{2}{3}-\dfrac{4}{9}\right)=\dfrac{-3}{4}\)
\(\dfrac{1}{3}x+\dfrac{2}{9}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{-35}{36}\)
\(\Leftrightarrow x=\dfrac{-35}{12}\)
Vậy ...
a) \(7-\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}=7\)
\(\Rightarrow x=\left(\sqrt{7}\right)^2\)
b) \(5\sqrt{x}+1=40\)
\(\Rightarrow5\sqrt{x}=39\)
\(\Rightarrow\sqrt{x}=7,8\)
\(\Rightarrow x=\left(\sqrt{7,8}\right)^2\)
c) \(\dfrac{5}{12}\sqrt{x}-\dfrac{1}{6}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{5}{12}\sqrt{x}=\dfrac{1}{2}\)
\(\Rightarrow\sqrt{x}=1,2\)
\(\Rightarrow x=\left(\sqrt{1,2}\right)^2\)
d) \(4x^2-1=0\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\Rightarrow x=0,5\\2x+1=0\Rightarrow x=-0,5\end{matrix}\right.\)
e) \(\sqrt{x+1}-2=0\)
\(\Rightarrow\sqrt{x+1}=2\)
\(\Rightarrow x+1=1,414\)
\(\Rightarrow x=0,414\)
f) \(2x^2+0,82=1\)
\(\Rightarrow2x^2=0,18\)
\(\Rightarrow x^2=0,09\)
\(\Rightarrow x=\pm0,3\)
g) Không có kết quả
Thêm đề: Tìm \(x\in Z\)
\(-\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\ge x\ge-4\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(\Rightarrow-\dfrac{2}{3}.\left(-\dfrac{11}{12}\right)\ge x\ge-\dfrac{13}{3}.\dfrac{1}{3}\)
\(\Rightarrow\dfrac{11}{18}\ge x\ge-\dfrac{13}{9}\)
Vì \(x\in Z\) nên \(x\in\left\{-1;0\right\}\)
Vậy...............
Chúc bạn học tốt!!!
Tìm x dễ thì tự làm nha:
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)-\left(\dfrac{x+2}{2002}+1\right)-\left(\dfrac{x+1}{2003}\right)=0\)\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
a,
\(\dfrac{5^x}{125}=5^4\\ 5^x:5^3=5^4\\ 5^x=5^4\cdot5^3\\ 5^x=5^7\\ \Rightarrow x=7\)
b,
\(\dfrac{3^x}{3}+3^{x-2}=4\\ 3^{x-1}+3^{x-2}=3^1+3^0\\ \Rightarrow x=2\)
c,
\(\left(x+\dfrac{2006}{2007}\right)^6=0\\ \Rightarrow x+\dfrac{2006}{2007}=0\\ x=0-\dfrac{2006}{2007}\\ x=\dfrac{-2006}{2007}\)
d,
\(\left(x-\dfrac{1}{5}\right)^3=\dfrac{8}{125}\\ \left(x-\dfrac{1}{5}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{1}{5}=\dfrac{2}{5}\\ x=\dfrac{2}{5}+\dfrac{1}{5}\\ x=\dfrac{3}{5}\)
e,
\(3^x+3^{x-2}=810\\ 3^x\left(1+3^2\right)=810\\ 3^x\cdot10=810\\ 3^x=810:10\\ 3^x=81\\ 3^x=3^4\\ \Rightarrow x=4\)
g,
\(5^{x+2}+5^{x+1}+5^x=19375\\ 5^x\left(5^2+5+1\right)=19375\\ 5^x\cdot31=19375\\ 5^x=19375:31\\ 5^x=625\\ 5^x=5^4\\ \Rightarrow x=4\)
a: \(\Leftrightarrow-\dfrac{2}{3}\cdot\dfrac{4-6-9}{12}\ge x\ge-\dfrac{13}{3}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow-\dfrac{2}{3}\cdot\dfrac{-11}{12}\ge x\ge\dfrac{-13}{3}\cdot\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{22}{36}\ge x\ge\dfrac{-13}{9}\)
mà x là số nguyên
nên \(x\in\left\{0;-1\right\}\)
b: \(\Leftrightarrow\dfrac{21}{100}+\dfrac{75}{100}-\dfrac{220}{100}>=2x-1>=-3-\dfrac{1}{2}+3+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{-124}{100}\ge2x-1\ge\dfrac{-3}{10}\)
\(\Leftrightarrow-\dfrac{124}{100}+1\ge2x>=\dfrac{-3}{10}+1\)
\(\Leftrightarrow\dfrac{-3}{25}\ge2x\ge\dfrac{7}{10}\)(vô lý)
=>x không có giá trị
c: \(\Leftrightarrow43+\dfrac{1}{2}-39-\dfrac{1}{5}\le-3x+4\le9+\dfrac{1}{5}+50+\dfrac{1}{7}\)
\(\Leftrightarrow3+\dfrac{3}{10}\le-3x+4\le59+\dfrac{12}{35}\)
\(\Leftrightarrow\dfrac{33}{10}-4\le-3x\le59+\dfrac{12}{35}-4\)
\(\Leftrightarrow\dfrac{-7}{10}\le-3x\le\dfrac{1937}{35}\)
\(\Leftrightarrow\dfrac{7}{30}\ge x\ge-\dfrac{1937}{105}\)
mà x là số nguyên
nên \(x\in\left\{0;-1;-2;...;-18\right\}\)
\(a,\dfrac{4,5}{x}=\dfrac{3}{5}\)
\(\Leftrightarrow3x=4,5.5\)
\(\Leftrightarrow3x=22,5\)
\(\Leftrightarrow x=7,5\)
Vậy ....
b, \(\dfrac{-x}{\dfrac{4}{5}}=\dfrac{\dfrac{-1}{5}}{x}\)
\(\Leftrightarrow\left(-x\right)x=\dfrac{4}{5}.\dfrac{-1}{5}\)
\(\Leftrightarrow\left(-x\right)x=\dfrac{-4}{25}\)
\(\Leftrightarrow\left(-x\right)x=\dfrac{-2}{5}.\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{2}{5}\)
Vậy ....
c, \(\dfrac{x-3}{2}=\dfrac{x}{5}\)
\(\Leftrightarrow\left(x-3\right)5=2x\)
\(\Leftrightarrow5x-15=2x\)
\(\Leftrightarrow5x-2x=15\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
Vậy .......
Áp dụng công thức ra nhé :
\(\dfrac{4,5}{x}=\dfrac{3}{5}\Rightarrow3x=22,5\Rightarrow x=7,5\)
Câu b : Ko bt tính từ dưới lên hay từ trên xuống nữa
Câu cờ :
\(\dfrac{x-3}{2}=\dfrac{x}{5}\Rightarrow\left(x-3\right).5=2x\)
\(\Rightarrow5x-15-2x=0\)
\(\Rightarrow3x-15=0\)
\(\Rightarrow x=5\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a/ \(VT=\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1=\left(1\right)\)
\(VP=\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b/ \(VT=\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)
\(VP=\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
c/ \(VT=\dfrac{2a-5b}{2c-5d}=\dfrac{2bk-5b}{2dk-5d}=\dfrac{b\left(2k-5\right)}{d\left(2k-5\right)}=\dfrac{b}{d}\left(1\right)\)
\(VP=\dfrac{3a+4b}{3c+4d}=\dfrac{3bk+4b}{3dk+4d}=\dfrac{b\left(3k+4\right)}{d\left(3k+4\right)}=\dfrac{b}{d}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{2a-5b}{2c-5đ}=\dfrac{3a+4b}{3c+4d}\)
d/ \(VT=\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{b^2-k^2}=\dfrac{k^2\left(b^2-d^2\right)}{b^2-d^2}=k^2\left(1\right)\)
\(VP=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)
\(\dfrac{MK}{MI}=\dfrac{2}{3}\Rightarrow MK=\dfrac{2}{3}MI\\ MK+KI=MI\\ \Leftrightarrow\dfrac{2}{3}MI+KI=MI\\ \Leftrightarrow KI=\dfrac{1}{3}MI\\ \dfrac{MK}{KI}=\dfrac{\dfrac{2}{3}MI}{\dfrac{1}{3}MI}=2\cdot\dfrac{\dfrac{1}{3}MI}{\dfrac{1}{3}MI}=2\)
=> Lại chọn C