a, 4*(3-x)^10=9(3-x)^8
b, (x-1)^x+3=(x-1)^x+1
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a.
\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
b.
\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)
c.
\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
d.
\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
e.
\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)
f.
\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)
a: \(x+6\dfrac{1}{8}=8\)
=>\(x+\dfrac{49}{8}=\dfrac{64}{8}\)
=>\(x=\dfrac{64}{8}-\dfrac{49}{8}=\dfrac{15}{8}\)
b: \(\dfrac{11}{2}\cdot x=\dfrac{1}{5}:\dfrac{1}{3}\)
=>\(x\cdot\dfrac{11}{2}=\dfrac{1}{5}\cdot3=\dfrac{3}{5}\)
=>\(x=\dfrac{3}{5}:\dfrac{11}{2}=\dfrac{3}{5}\cdot\dfrac{2}{11}=\dfrac{6}{55}\)
c: \(x\cdot\dfrac{3}{5}+\dfrac{2}{5}\cdot x=\dfrac{4}{9}+\dfrac{1}{3}\)
=>\(x\left(\dfrac{3}{5}+\dfrac{2}{5}\right)=\dfrac{4}{9}+\dfrac{3}{9}\)
=>\(x\cdot1=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}\)
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
a) X + 1/4 = 5/8
b) X - 3/5 = 1/10
c) X x 2/7 = 6/11
d) X : 3/2 = 1/4
đ) 7/4 - X = 5/7
e) 5/4 : x = 1/8
\(a,x+\dfrac{1}{4}=\dfrac{5}{8}\)
\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{2}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
\(b,x-\dfrac{3}{5}=\dfrac{1}{10}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{6}{10}\)
\(\Leftrightarrow x=\dfrac{7}{10}\)
\(c,x\times\dfrac{2}{7}=\dfrac{6}{11}\)
\(\Leftrightarrow x=\dfrac{6}{11}:\dfrac{2}{7}\)
\(\Leftrightarrow x=\dfrac{6}{11}\times\dfrac{7}{2}\)
\(\Leftrightarrow x=\dfrac{42}{22}\)
\(\Leftrightarrow x=\dfrac{21}{11}\)
\(d,x:\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}\times\dfrac{3}{2}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
\(đ,\dfrac{7}{4}-x=\dfrac{5}{7}\)
\(\Leftrightarrow x=\dfrac{7}{4}-\dfrac{5}{7}\)
\(\Leftrightarrow x=\dfrac{49}{28}-\dfrac{20}{28}\)
\(\Leftrightarrow x=\dfrac{29}{28}\)
\(e,\dfrac{5}{4}:x=\dfrac{1}{8}\)
\(x=\dfrac{5}{4}:\dfrac{1}{8}\)
\(\Leftrightarrow x=\dfrac{5}{4}\times8\)
\(\Leftrightarrow x=\dfrac{40}{4}\)
\(\Leftrightarrow x=10\)
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\Rightarrow x=-2\\ b,\Rightarrow\left(x-2021\right)\left(2022x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{2022}\end{matrix}\right.\\ c,\Rightarrow\left(x^2-9\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(2x+7\right)=0\\ \Rightarrow\left(x-3\right)\left(x+3-2x-7\right)=0\\ \Rightarrow\left(x-3\right)\left(-4-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
`x/8 = 3/4 +(-5/8)`
`=>x/8 = 6/8 +(-5/8)`
`=>x/8 = 1/8`
`=>x=1`
`-----`
`x/12 =3/4 +(-2/3)`
`=>x/12 = 9/12 + (-8/12)`
`=> x/12=1/12`
`=>x=1`
`----`
`1+11/13=24/x`
`=> 13/13 +11/13=24/x`
`=> 24/13 =24/x`
`=>x=13`
`----`
`x/6 -3/4=1/12`
`=>x/6 = 1/12 +3/4`
`=>x/6 = 1/12 + 9/12`
`=>x/6 = 10/12`
`=>x/6= 5/6`
`=>x=5`
`4.(3-x)^10=9(3-x)^8`
`=>(3-x)^{8}(4(3-x)^2-9)=0`
`=>` $\left[ \begin{array}{l}3-x=0\\(3-x)^2=\dfrac{9}{4}\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=3\\3-x=\dfrac{3}{2}\\3-x=-\dfrac{3}{2}\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=3\\x=\dfrac{3}{2}\\x=\dfrac{9}{2}\end{array} \right.$
Vậy `x=3` hoặc `x=3/2` hoặc `x=9/2`
`b,(x-1)^{x+3}=(x-1)^{x+1}`
`=>(x-1)^{x+1}[(x-1)^2-1]=0`
`=>` $\left[ \begin{array}{l}x-=1\\x-=-1\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=2\\x=1\\x=0\end{array} \right.$
Vậy `x=0` hoặc `x=1` hoặc `x=2`
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