Tìm x biết:
a)\((\frac{-5}{9})^{10}:x=(\frac{-5}{9})^8\)
giúp vs ạ
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a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
a)
\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)
Vậy \(x = \frac{{ - 3}}{2}\).
b)
\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
c)
\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)
Vậy \(x = \frac{4}{5}\)
d)
\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)
Vậy \(x = \frac{{ - 2}}{5}\).
Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.
a)
\(\begin{array}{l}x - \left( { - \frac{7}{9}} \right) = - \frac{5}{6}\\x + \frac{7}{9} = - \frac{5}{6}\\x = - \frac{5}{6} - \frac{7}{9}\\x = - \frac{{15}}{{18}} - \frac{{14}}{{18}}\\x = \frac{{ - 29}}{{18}}\end{array}\)
Vậy \(x = \frac{{ - 29}}{{18}}\).
b)
\(\begin{array}{l}\frac{{15}}{{ - 4}} - x = 0,3\\x = \frac{{15}}{{ - 4}} - 0,3\\x = - 3,75 - 0,3\\x = - 4,05\end{array}\)
Vậy \(x = - 4,05\).
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
a)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
\(\frac{3}{5}x-\frac{13}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{13}{9}:13:\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{1}{9}:\left(\frac{21}{315}+\frac{9}{315}+\frac{5}{315}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{1}{9}:\frac{35}{315}=-10\)
<=> \(\frac{3}{5}x-\frac{1}{9}:\frac{1}{9}=-10\)
<=> \(\frac{3}{5}x-1=-10\)
<=> \(\frac{3}{5}x=-9\)
<=> \(x=-15\)
Vậy x = -15.
\(\frac{3}{5}x-1\frac{4}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}=-10\right)\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\right)\right]=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{3.5}+\frac{2}{.57}+\frac{2}{7.9}\right)\right]=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{1}{3}-\frac{1}{9}\right)\right]=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{2}.\frac{2}{9}\right)=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\frac{26}{18}=-10\)
\(\Leftrightarrow\frac{3}{5}x-1=-10\)
\(\Leftrightarrow\frac{3}{5}x=-10+1\)
\(\Leftrightarrow\frac{3}{5}x=-9\)
\(\Rightarrow x=-9:\frac{3}{5}\)
\(\Rightarrow x=-15\)
Vậy \(x=-15\)
4^5=2^10
9^4=3^8
2*6^9=2^10*3^9
thì cái tử sẽ đc:
2^10*(-3)
mẫu e phân tích tt
\(A=\frac{5}{x+3}-\frac{2}{3-x}-\frac{3x^2-2x-9}{x^2-9}\)
a) ĐKXĐ: \(\hept{\begin{cases}x+3\ne0\\3-x\ne0\\x^2-9\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne3\\x\ne3;x\ne-3\end{cases}}}\)
Vậy ĐKXĐ: x khác -3; x khác 3 ( b vào tcn của mìnk để thấy chi tiết)
Rút gọn:
\(A=\frac{5}{x+3}-\frac{2}{3-x}-\frac{3x^2-2x-9}{x^2-9}\)
\(\Leftrightarrow A=\frac{5}{x+3}+\frac{2}{x-3}-\frac{3x^2-2x-9}{\left(x-3\right)\left(x+3\right)}\) MTC: (x-3)(x+3)
\(\Leftrightarrow A=\frac{5\left(x-3\right)+2\left(x+3\right)-\left(3x^2-2x-9\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{9x-3x^2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{3x\left(3-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-3x}{x+3}\)
Vậy A=-3x/x+3 với x khác 3 và x khác -3
b) |x-2|=1
Bỏ dấu gt tuyệt đối ta có 2 TH: (đối chiếu đkxđ)
* x-2=1=> x=1+2=>x=3 (o t/m)
*x-2=-1=>x=-1+2=>x=1 (tm)
Thay x=1 vào phân thức A rút gọn ta có:
\(A=\frac{-3x}{x+3}=\frac{-3.1}{1+3}=\frac{-3}{4}\)
Vậy A=-3/4 khi x=1
c) Để A có gt nguyên => A thuộc Z
=> \(A=\frac{-3x}{x+3}\in Z\)
Ta có: -3x chia hết x+3
=> -3(x-3)-9 chia hết x+3
=> -9 chia hết cho x+3
=> x+3 thược Ư(-9)={1;-1;9;-9;3;-3)
Lập bảng thay vào hoặc o cần cx được
x+3 | 1 | -1 | 9 | -9 | 3 | -3 |
x | -2(tm) | -4(tm) | 6(tm) | -12(tm) | 0(tm) | -6(tm) |
Vậy...
Lời giải: Giải phương trình với tập xác định
Tập xác định của phương trình
\(x\in\infty-\infty\)
\(\frac{19x+67}{90}=\frac{15x+83}{56}\Rightarrow\left(19x=67\right)56=90\left(15x+83\right)\)
Kết quả : \(-13\)
kq đúng nhưng mk k biết mấy cái phương trình đó vì mk mới lớp 7
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
Ta có: \(\left(-\frac59\right)^{10}:x=\left(-\frac59\right)^8\)
=>\(x=\left(-\frac59\right)^{10}:\left(-\frac59\right)^8=\left(-\frac59\right)^2=\frac{25}{81}\)