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a. \(\dfrac{-39}{7}:x=26\)
x = \(\dfrac{-39}{7}:26\)
x = \(\dfrac{-3}{14}\)
b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)
x = \(\dfrac{7}{4}.\dfrac{13}{5}\)
x = \(\dfrac{91}{20}\)
c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)
x = \(\dfrac{-11}{10}\)
d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)
x = \(\dfrac{9}{4}+\dfrac{3}{4}\)
x = 3
e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)
x = \(\dfrac{7}{8}:\dfrac{14}{3}\)
x = \(\dfrac{3}{16}\)
f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)
x = \(\dfrac{13}{3}.\dfrac{8}{3}\)
x = \(\dfrac{104}{9}\)
g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)
x = 0
chúc bạn học tốt 
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
a) \(\dfrac{x}{48}=-\dfrac{4}{7}\Rightarrow x=-\dfrac{192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\Rightarrow x+\dfrac{4}{5}=1\)
\(\Rightarrow x=\dfrac{1}{5}\)
c) \(2\left|x-1\right|^2=72\Rightarrow\left|x-1\right|^2=36\)
\(\Rightarrow\left|x-1\right|=6\)
TH1: x - 1 = -6 => x = -5
TH2: x - 1 = 6 => x = 7
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\Rightarrow x=2\)
f) | x - 2 | = 1 + 4 = 5
TH1: x - 2 = -5 => x = -3
TH2: x - 2 = 5 => x = 7
a) \(\dfrac{x}{48}=\dfrac{-4}{7}\)
⇒ x.7=48.(-4)
7x = -192
x=\(\dfrac{-192}{7}\) Vậy x=\(\dfrac{-192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\)
\(\left(x+\dfrac{4}{5}\right)=\dfrac{3}{5}+\dfrac{2}{5}\)
\(x+\dfrac{4}{5}=1\)
\(x=1-\dfrac{4}{5}\)
\(x=\dfrac{1}{5}\)
c) chưa từng gặp dạng với giá trị tuyệt đối sory
d) \(\dfrac{1}{6}x-\dfrac{2}{3}=2\)
\(\dfrac{1}{6}x=2+\dfrac{2}{3}\)
\(\dfrac{1}{6}x=\dfrac{8}{3}\)
\(x=\dfrac{8}{3}:\dfrac{1}{6}\)
\(x=16\)
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\)
=> x.5 = 4.2,5
5x=10
x=10:5
x=2
f) |x-2|-4=1
|x-2|=1+4
|x-2|=5
=>\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=5+2\\x=-5+2\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
đôi khi cũng có sai sót , hãy xem lại thật kĩ
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
\(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{98}+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{97}+\left(\dfrac{1}{2}\right)^{98}\)
\(\Rightarrow2B-B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(B=1-\left(\dfrac{1}{2}\right)^{99}\)
\(2,\)
\(a,\dfrac{45^{10}.2^{10}}{75^{15}}\)
\(=\dfrac{5^{10}.9^{10}.2^{10}}{25^{15}.3^{15}}\)
\(=\dfrac{5^{10}.3^{20}.2^{10}}{5^{30}.3^{15}}\)
\(=\dfrac{5^{10}.3^{15}.\left(3^5.2^{10}\right)}{5^{10}.3^{15}.\left(5^{20}\right)}\)
\(=\dfrac{3^5.2^{10}}{5^{20}}\)
\(b,\dfrac{2^{15}.9^4}{6^3.8^3}\)
\(=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
\(c,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{4^{10}.2^{10}+4^{10}}{4^4.2^4+4^4.4^7}=\dfrac{4^4.\left(4^6.2^{10}+4^6\right)}{4^4.\left(2^4+4^7\right)}\)
\(=\dfrac{4^{11}+4^6}{4^8.4^7}=\dfrac{4^6.\left(4^5+1\right)}{4^6.\left(4^2-4\right)}=\dfrac{1024+1}{16-4}=\dfrac{1025}{12}\)
\(d,\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
\(3,\)
\(a,\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x+4=\dfrac{1}{2}\\2x+4=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}-4=\dfrac{-7}{2}\\2x=\dfrac{-1}{2}-4=\dfrac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-7}{4};\dfrac{-9}{4}\right\}\)
\(b,\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2=\left(-6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6+3=9\\2x=-6+3=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{9}{2};\dfrac{-3}{2}\right\}\)
\(c,5^{x+2}=628\)
\(5^{x+2}=5^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=4-2=2\)
Vậy \(x=2\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
Bài 1:
B= \(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{99}\)
2B= \(2.[\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}]\)
2B= \(1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{98}\)
⇒2B-B= \(1-\left(\dfrac{1}{2}\right)^{99}\)
B= 1
Vậy B=1
Bài 2:
a, \(\dfrac{45^{10}.2^{10}}{75^{15}}\)= \(\dfrac{\left(3^2.5\right)^{10}.2^{10}}{\left(3.5^2\right)^{15}}=\dfrac{3^{20}.5^{10}.2^{10}}{3^{15}.5^{30}}=\dfrac{3^5.2^{10}}{5^{20}}\)
b, \(\dfrac{2^{15}.9^4}{6^3.8^3}=\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\dfrac{2^{15}.3^8}{2^3.3^3.2^9}=\dfrac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5\)
c,\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2.4\right)^{10}+4^{10}}{\left(2.4\right)^4+4^{11}}=\dfrac{2^{10}.4^{10}+4^{10}}{2^4.4^4+4^{11}}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6+4^6.4^5}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(4^5+1\right)}=\dfrac{4^{10}.\left(2^{10}+1\right)}{4^6.\left(2^{10}+1\right)}=4^4=256\)
d, \(\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
Bài 3:
a, \(\left(2x+4\right)^2=\dfrac{1}{4}\)
\(\left(2x+4\right)^2=\left(\dfrac{1}{2}\right)^2\)
\(2x+4=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-4\)
\(2x=-\dfrac{7}{2}\)
\(x=-\dfrac{7}{2}:2\)
\(x=-\dfrac{7}{2}.\dfrac{1}{2}\)
\(x=-\dfrac{7}{4}\)
b, \(\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2\)
\(2x-3=6\)
\(2x=9\)
\(x=\dfrac{9}{2}\)
c, \(5^{x+2}=625\)
\(5^{x+2}=5^4\)
\(x+2=4\)
\(x=2\)
a) \(2\left(4x-30\right)-3\left(x+5\right)+4\left(x-10\right)=5\left(x+2\right)\)
\(\Leftrightarrow8x-60-3x+15+4x-40=5x+10\)
\(\Leftrightarrow9x-35=5x+10\)
\(\Leftrightarrow9x-5x=10+35\)
\(\Leftrightarrow4x=45\)
\(\Leftrightarrow x=\dfrac{45}{4}=11,25\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\left(6x+1\right)\)
\(\Leftrightarrow\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=4x+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{31}{60}+x=4x+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{31}{60}-\dfrac{2}{3}=4x-x\)
\(\Leftrightarrow3x=\dfrac{1}{60}\)
\(\Leftrightarrow x=\dfrac{1}{180}\)
c) \(\dfrac{7}{3}-\left(2x-\dfrac{1}{3}\right)=\left(-2\dfrac{1}{6}+1\dfrac{1}{2}\right):0,25\)
\(\Leftrightarrow\dfrac{7}{3}-2x+\dfrac{1}{3}=-1\dfrac{2}{3}:\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-5}{3}.4\)
\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-20}{3}\)
\(\Leftrightarrow2x=\dfrac{8}{3}+\dfrac{20}{3}\)
\(\Leftrightarrow2x=\dfrac{28}{3}\)
\(\Leftrightarrow x=4\dfrac{2}{3}\)
d) \(0,75+\dfrac{5}{9}:x=5\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{4}+\dfrac{5}{9}:x=\dfrac{11}{2}\)
\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{11}{2}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{19}{4}\)
\(\Leftrightarrow x=\dfrac{5}{9}:\dfrac{19}{4}\)
\(\Leftrightarrow x=\dfrac{20}{171}\)
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
b) \(\left|x+\dfrac{3}{4}\right|-5=-2\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|=-2+5\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{15}{4}\end{matrix}\right.\)
a) \(\left|1-x\right|+0,37=3\)
\(\Rightarrow\left|1-x\right|=2,63\)
\(\Rightarrow\left[{}\begin{matrix}1-x=2,63\\1-x=-2,63\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1,63\\x=3,63\end{matrix}\right.\)