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Bài làm:
Ta có: \(3^{2x+2}=9^{x+3}\)
\(\Leftrightarrow\left(3^2\right)^{x+1}=9^{x+3}\)
\(\Leftrightarrow9^{x+1}=9^{x+3}\)
\(\Rightarrow x+1=x+3\)
\(\Rightarrow0x=-2\) (vô lý)
Vậy không tồn tại x thỏa mãn PT
<=> |x+2| = 13
<=> \(\orbr{\begin{cases}x+2=13\\x+2=-13\end{cases}\Rightarrow}\orbr{\begin{cases}x=11\\x=-15\end{cases}}\)
Vậy.........
hok tốt
..........
\(|x+2|=12+\left(-3\right)+\left|-4\right|\)
\(|x+2|=12-3+4\)
\(\left|x+2\right|=13\)
\(\Rightarrow x\in\left\{-15;11\right\}\)
a) x.(\(\dfrac{6}{7}\)+\(\dfrac{5}{6}\))=\(\dfrac{3}{4}\)
x.\(\dfrac{71}{42}\)=\(\dfrac{3}{4}\)
x=\(\dfrac{3}{4}\):\(\dfrac{71}{42}\)
x=\(\dfrac{63}{142}\)
a.\(\dfrac{6}{7}x+\dfrac{5}{6}x=\dfrac{3}{4}\)
\(x.\left(\dfrac{6}{7}+\dfrac{5}{6}\right)=\dfrac{3}{4}\)
\(x.\dfrac{71}{42}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{71}{42}\)
\(x=\dfrac{63}{142}\)
b\(\dfrac{5}{4}-\dfrac{3}{5}:x=1\dfrac{1}{3}\)
\(\dfrac{3}{5}:x=\dfrac{5}{4}-1\dfrac{1}{3}\)
\(\dfrac{3}{5}:x=\dfrac{-1}{12}\)
\(x=\dfrac{3}{5}:\dfrac{-1}{12}\)
\(x=\dfrac{-36}{5}\)
c. \(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right):3\dfrac{1}{2}=0,5\)
\(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right)=0,5:3\dfrac{1}{2}\)
\(\dfrac{4}{7}x-\dfrac{1}{3}=\dfrac{1}{7}\)
\(\dfrac{4}{7}x=\dfrac{1}{7}+\dfrac{1}{3}\)
\(\dfrac{4}{7}x=\dfrac{10}{21}\)
\(x=\dfrac{10}{21}:\dfrac{4}{7}\)
\(x=\dfrac{5}{6}\)
d.\(\dfrac{4}{5}-\dfrac{2}{3}x=1\dfrac{1}{4}+2,5x\)
\(\dfrac{4}{5}-\left(\dfrac{2}{3}x-2,5x\right)=1\dfrac{1}{4}\)
\(\dfrac{4}{5}-\dfrac{-11}{6}x=1\dfrac{1}{4}\)
\(\dfrac{-11}{6}x=\dfrac{4}{5}-1\dfrac{1}{4}\)
\(\dfrac{-11}{6}x=\dfrac{-9}{20}\)
\(x=\dfrac{-9}{20}:\dfrac{-11}{6}\)
\(x=\dfrac{27}{110}\)
có sai sót j xin bn thông cảm !
a)\(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}+\dfrac{5}{6}\)
\(\dfrac{2}{3}x=\dfrac{25}{12}\)
\(x=\dfrac{25}{12}:\dfrac{2}{3}\)
=>\(x=\dfrac{25}{8}\)
a) \(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\) b) \(2\dfrac{1}{3}-\dfrac{4}{5}:x=0,2\)
\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\) \(\dfrac{7}{3}-\dfrac{4}{5}:x=\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{5}{4}-\dfrac{5}{6}\) \(\dfrac{4}{5}:x=\dfrac{7}{3}-\dfrac{1}{5}\)
\(\dfrac{2}{3}x=\dfrac{30}{24}-\dfrac{20}{24}\) \(\dfrac{4}{5}:x=\dfrac{35}{15}-\dfrac{3}{15}\)
\(\dfrac{2}{3}x=\dfrac{5}{12}\) \(\dfrac{4}{5}:x=\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{2}{3}\) \(x=\dfrac{4}{5}:\dfrac{32}{15}\)
\(x=\dfrac{5}{12}:\dfrac{8}{12}\) \(x=\dfrac{4}{5}.\dfrac{15}{32}\)
\(x=\dfrac{5}{12}.\dfrac{12}{8}=\dfrac{5}{8}\) \(x=\dfrac{4.15}{5.32}\)
\(x=\dfrac{1.3}{1.8}=\dfrac{3}{8}\)
d)\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\dfrac{-8}{27}\)
\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\left(\dfrac{-2}{3}\right)^3\)
\(\Rightarrow\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=\dfrac{4}{3}-\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{1}{4}x=2\)
\(\Rightarrow x=2:\dfrac{1}{4}\)
\(\Rightarrow x=2.4=8\)
a) \(\frac{x-1}{6}=\frac{2x+3}{7}\)
\(\Leftrightarrow7\left(x-1\right)=6\left(2x+3\right)\)
\(\Leftrightarrow7x-7=12x+18\)
\(\Leftrightarrow5x+18=-7\)
\(\Leftrightarrow5x=-25\)
\(\Leftrightarrow x=-5\)
b) \(\left(2x^2-\frac{1}{2}x\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{2}\right)\left(x^2+1\right)=0\)
Vì \(x^2+1>0\)nên \(\orbr{\begin{cases}x=0\\2x-\frac{1}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)
3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
a.\(\dfrac{-4}{5}-\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{2}{7}\)
\(\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{-4}{5}-\dfrac{2}{7}=\dfrac{-38}{35}\)
\(\dfrac{2}{3}x=\dfrac{-38}{35}-1\dfrac{1}{4}\)
\(\dfrac{2}{3}x=\dfrac{-327}{140}\Rightarrow x=\dfrac{-327}{140}:\dfrac{2}{3}=\dfrac{-981}{280}\)
Vậy \(x=\dfrac{-981}{280}\)
b. \(\dfrac{5}{6}+\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}\)
\(\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}-\dfrac{5}{6}=\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{3}{4}-\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{9}{4}\Rightarrow x=\dfrac{1}{2}:\dfrac{9}{4}=\dfrac{2}{9}\)
Vậy \(x=\dfrac{2}{9}\)
c. \(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right):\dfrac{3}{4}=0,7\)
\(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right)=0,7.\dfrac{3}{4}=\dfrac{21}{40}\)
\(\dfrac{4}{5}x=\dfrac{21}{40}+1\dfrac{1}{3}=\dfrac{223}{120}\)
\(\Rightarrow x=\dfrac{223}{120}:\dfrac{4}{5}=\dfrac{223}{96}\)
Vậy \(x=\dfrac{223}{96}\)
d. \(\dfrac{5}{6}-\dfrac{3}{4}x=1\dfrac{1}{3}+0,5x\)
\(0,5x+\dfrac{3}{4}x=\dfrac{5}{6}-1\dfrac{1}{3}\)
\(\dfrac{5}{4}x=\dfrac{-1}{2}\Rightarrow x=\dfrac{-1}{2}:\dfrac{5}{4}=\dfrac{-2}{5}\)
Vậy \(x=\dfrac{-2}{5}\)
a.=>-3\(⋮\) x-1
x-1 thuộc ước của -3
x-1=1=>x=1+1=
x-1=-1=>....
x-1=3=>..
x-1=-3=>......
b. tương tự câu a
c.\(\frac{3x+7}{x-1}=\frac{3x-3+10}{x-1}=\frac{3\left(x-1\right)}{x-1}=\frac{10}{x-1}\)
Tự tính tiếp nha
d.chịu
a) Để \(\frac{-3}{x-1}\) nguyên <=> x -1 \(\varepsilon\) Ư(-3)
ta có Ư(-3) = {-3 ; 3 ; 1; -1 }.
Với x -1 = 1 <=> x=2
Với x-1 =-1 <=> x= 0
Với x-1 =3 <=> x=4
Với x-1 =-3 <=> x=-2
Vậy.......
ý b bạn làm tương tự nhé có j hỏi mk thêm mk sẽ hướng dẫn ý c và d cho đỡ tồn thời gian
c) \(\frac{3x+7}{x-1}\)
=\(\frac{3x-3+10}{X-1}=\frac{3\left(x-1\right)+10}{x-1}\)
=\(\frac{3\left(x-1\right)}{x-1}+\frac{10}{x-1}\)
= 3 +\(\frac{10}{x-1}\)
Để \(\frac{3x+7}{x-1}\) nguyên <=> x -1\(\varepsilon\) Ư(10)
ta có Ư(10) ={-1; 1 ; -2 ; 2 ; 5 ; -5 , 10 ; -10}.
Với x -1 = -1 <=> x=0
Với x -1 = 1<=> x= 2
Với x-1=-2 <=> x= -1
Với x-1=2 <=> x= 3
Với x-1 =5 <=> x=5
Với x-1=-5<=>x=-4
Với x-1= 10<=>x=11
Với x-1=-10<=>x=-9
VẬY ...................................
D) \(\frac{4x-1}{3-x}\)
=\(\frac{4x-12+11}{3-x}\)
=\(\frac{4\left(x-3\right)+11}{3-x}\)
=\(\frac{4\left(x-3\right)}{-\left(x-3\right)}+\frac{11}{3-x}\)
= -4+ \(\frac{11}{3-x}\)
Để \(\frac{4x-1}{3-x}\) nguyên <=> 3-x\(\varepsilon\) Ư(11)={-1 ; 1 ;-11 ;11 }.
Với 3 -x =-1 <=> x=4
Với 3 -x =1 <=> x=2
Với 3 -x = -11 <=> x=14
Với 3 -x = 11 <=> x = -8
VẬY ........................
ĐÂY LÀ CACH GIẢI CHI TIẾT NHẤT ĐẤY . CHÚC BẠN NGÀY CÀNG HỌC GIỎI. NHỚ CHO MK NHÉ
\(\dfrac{2x-6}{3}=\dfrac{x+3}{4}\)
=>4(2x-6)=3(x+3)
=>8x-24=3x+9
=>5x=33
=>\(x=\dfrac{33}{5}\)
=>( 2x-6).4=(x+3).3=>8x-24=3x+9=>8x+3x=24+9=>11x=33=>x=11.Vậy x=11