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21 tháng 8 2021

a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)

\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)

\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)

\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))

\(a,\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\frac{11}{15}x=\frac{2}{5}\)

\(x=\frac{6}{11}\)

b,\(\left(2x-3\right).\left(6-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

Vậy

25 tháng 10 2023

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

20 tháng 2 2021

cái chỗ math processing error kia là \(3\left(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}\right)+\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)\ge\dfrac{985}{108}\)

23 tháng 11 2021

345x2-107x2=690-214=476 hoặc bạn nhóm 2 ra ngoài

425x3-17x3=1275-51=1124 hoặc nhóm 3 ra ngoài

108x6-108x2=648-216=432 hoặc nhóm 108 ra

:))))))))))) dễ mà

17 tháng 11 2023

\(\dfrac{1}{x-3}-\dfrac{1}{x}=\dfrac{x-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{x-x+3}{x\left(x-3\right)}=\dfrac{3}{x\left(x-3\right)}\)

\(B=\dfrac{1}{x^2-3x}+\dfrac{1}{x^2-9x+18}+\dfrac{1}{x^2-15x+54}+\dfrac{1}{x^2-21x+108}\)

\(=\dfrac{1}{x\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-6\right)}+\dfrac{1}{\left(x-6\right)\left(x-9\right)}+\dfrac{1}{\left(x-9\right)\left(x-12\right)}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{x\left(x-3\right)}+\dfrac{3}{\left(x-3\right)\left(x-6\right)}+\dfrac{3}{\left(x-6\right)\left(x-9\right)}+\dfrac{3}{\left(x-9\right)\left(x-12\right)}\right)\)

\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-6}-\dfrac{1}{x-6}+\dfrac{1}{x-9}-\dfrac{1}{x-9}+\dfrac{1}{x-12}\right)\)

\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-12}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-\left(x-12\right)+x}{x\left(x-12\right)}\)

\(=\dfrac{4}{x\left(x-12\right)}\)

25 tháng 8 2021

1) \(5^{x+1}-5^x=20\Leftrightarrow5^x\left(5-1\right)=20\Leftrightarrow5^x=5\Leftrightarrow x=1\)

2) \(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\Leftrightarrow2^x=32\Leftrightarrow x=5\)

3) \(4^{2x+1}+4^{2x}=80\Leftrightarrow4^{2x}\left(4+1\right)=80\Leftrightarrow16^x=16\Leftrightarrow x=1\)

4) \(3^{2x+2}+3^{2x+1}=108\Leftrightarrow3^{2x}\left(3^2+3\right)=108\Leftrightarrow9^x=9\Leftrightarrow x=1\)

5) \(7^{x+3}-7^{x+1}=16464\Leftrightarrow7^x\left(7^3-7\right)=16464\Leftrightarrow7^x=49\Leftrightarrow x=2\)

25 tháng 8 2021

cảm ơn bn nha,