(3x + 21)3= \(\dfrac{27}{8}\)
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a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
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d: \(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8=12x^2-12x-8\)
\(\Leftrightarrow12x^2+16=12x^2-12x-8\)
=>-12x=24
hay x=-2
e: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(x-5\right)^2\)
\(\Leftrightarrow x^2+7x+10-12x+9=x^2-10x+25\)
=>-5x+19=-10x+25
=>5x=6
hay x=6/5
f: \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
=>x-105=0
hay x=105
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
a: \(\dfrac{3}{9}\times\dfrac{5}{4}=\dfrac{5}{4}\times\dfrac{1}{3}=\dfrac{5\times1}{4\times3}=\dfrac{5}{12}\)
b: \(\dfrac{10}{15}\times\dfrac{3}{5}=\dfrac{3}{5}\times\dfrac{2}{3}=\dfrac{3\times2}{5\times3}=\dfrac{2}{5}\)
c: \(\dfrac{5}{8}\times\dfrac{4}{12}=\dfrac{5}{8}\times\dfrac{1}{3}=\dfrac{5}{3\times8}=\dfrac{5}{24}\)
d: \(\dfrac{9}{27}\times\dfrac{3}{21}=\dfrac{1}{7}\times\dfrac{1}{3}=\dfrac{1\times1}{7\times3}=\dfrac{1}{21}\)
a, \(\Leftrightarrow3x^2-3+5=3x^2+2x-3x-2\)
\(\Leftrightarrow3x^2-3x-2x+3x=-2+3-5\)
<=>x=-4
b, \(\Leftrightarrow\dfrac{x+4}{5}-\dfrac{5x}{5}+\dfrac{20}{5}=\dfrac{2x}{6}-\dfrac{3\left(x-2\right)}{6}\)
\(\Leftrightarrow\dfrac{x+4-5x+20}{5}=\dfrac{2x-3x+6}{6}\)
\(\Leftrightarrow\dfrac{6\left(-4x+24\right)}{30}=\dfrac{5\left(-x+6\right)}{30}\)
<=>-24x+144=-5x+30
<=>-5x+24x=144-30
<=>19x=114
<=>x=6
1)
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)
\(\Leftrightarrow x=105\)
b)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)
\(\Leftrightarrow50-x=0\)
\(\Leftrightarrow x=50\)
2)
\(\left(5x+1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)
\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)
\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)
\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
\(=\dfrac{3^{30}}{3^9\cdot7}+\dfrac{2}{3}-\dfrac{11}{7}=\dfrac{3^{21}-11}{7}+\dfrac{2}{3}=\dfrac{3^{22}-33+14}{21}=\dfrac{3^{22}-19}{21}\)
(3x+21)3=\(\dfrac{27}{8}\)
\(\Rightarrow\)(3x+21)3=\(\left(\dfrac{3}{2}\right)^3\)
\(\Rightarrow\)3x+21=\(\dfrac{3}{2}\)
\(\Rightarrow\)3x =\(\dfrac{3}{2}\)-21=\(\dfrac{-39}{2}\)
\(\Rightarrow\)x =\(\dfrac{-39}{2}\):3
\(\Rightarrow\) =\(\dfrac{-13}{2}\)
Vậy x =\(\dfrac{-13}{2}\)
\(\left(3x+21\right)^3=\dfrac{27}{8}\)
\(\Rightarrow3x+3.7=\dfrac{1}{2}.3\)
\(\Rightarrow x+7=\dfrac{1}{2}\)
\(\Rightarrow x=-7\dfrac{1}{2}\)