tìm x biết:
(3x + 4)3 = (9x + 8)2
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\(\left(3x+4\right)^3=\left(9x-8\right)\left(3x^2-8\right)\)
\(27x^3+108x^2+144x+64=27x^3-72x-24x^2+64\)
\(27x^3-27x^3+108x^2+24x^2+144x+72x=64-64=0\)
\(132x^2+216x=0\)
\(x\left(132x+216\right)=0\)
\(\Rightarrow x=\hept{\begin{cases}0\\\frac{216}{132}=\frac{18}{11}\end{cases}}\)
\(\Rightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\Rightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)
\(\Rightarrow-\frac{3}{2}x=-\frac{23}{4}\)
\(\Rightarrow x=\frac{23}{4}:\frac{3}{2}=\frac{23}{6}\)
\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+\frac{11}{3}\)
\(\Rightarrow\frac{5}{9}x=\frac{13}{3}\)
\(\Rightarrow x=\frac{13}{3}:\frac{5}{9}=\frac{39}{5}\)
\(\Rightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\Rightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)
\(\Rightarrow-\frac{3}{2}x=-\frac{23}{4}\)
\(\Rightarrow x=\frac{23}{4}:\frac{3}{2}=\frac{23}{6}\)
\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+\frac{11}{3}\)
\(\Rightarrow\frac{5}{9}x=\frac{13}{3}\)
\(\Rightarrow x=\frac{13}{3}:\frac{5}{9}=\frac{39}{5}\)
a)
<=> 3x - 3 + x - 2 = 2x - 2 - x + 1
<=> 3x + x - 2x + x = -2 + 1 + 3 + 2
<=> 3x = 4
<=> x = 4/3
Các câu sau làm tương tự
\(\left(3x-3\right)+\left(x-2\right)=\left(2x-2\right)-\left(x-1\right)\)
<=> \(3x-3+x-2=2x-2-x+1\)
<=> \(4x-5=x-1\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
Vậy....
(1-3x2)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2
⇒1-3x2-(9x2+x-18x-2)=9x2-16-9(x2+6x+9)
⇒1-3x2-(9x2-17x-2)= -56x-97
⇒1-3x2-9x2+17x+2=-56x-97
⇒3-12x2+17x=-56x-97
⇒3-12x2+17x+56x+97=0
⇒-12x2+73x+100=0
⇒-(12x2-73x-100)=0
b) ta có \(\left(4x-5\right)^3-\left(2x+5\right)\left(16x^2-25\right)=0\)
\(\left(4x-5\right)^3-\left(2x+5\right)\left(4x+5\right)\left(4x-5\right)=0\)
\(\left(4x-5\right)\left[\left(4x-5\right)^2-\left(2x+5\right)\left(4x+5\right)\right]=0\)
\(\left(4x-5\right)\left(16x^2-40x+5^2-8x^2-10x-20x-5^2\right)=0\)
\(\left(4x-5\right)\left(8x^2-70x\right)=0\)
=> \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}4x=5\\x\left(8x-70\right)=0\end{cases}< =>}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}\) \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}< =>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}}\)
\(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{35}{4}\end{cases}}\) Vậy \(\orbr{\begin{cases}x=\frac{5}{4}\\x=0\end{cases}}\) hoặc \(x=\frac{35}{4}\)
a) \(\left(3x-4\right)^3=\left(9x-8\right)\left(3x^2-8\right)\)
\(\Leftrightarrow27x^3+108x^2+144x+64=27x^3-72x-24x^2+64\)
\(\Leftrightarrow27x^3+108x^2+144x+64-27x^3+72x+24x^2-64=0\)
\(\Leftrightarrow132x^2+216x=0\)
\(\Leftrightarrow12x\left(11x+18\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\11x+18=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-\frac{18}{11}\end{array}\right.\)
\(\left(3x+4\right)^3=\left(9x+8\right)^2\)
\(\Leftrightarrow\)\(27x^3+108x^2+144x+64=81x^2+144x+64\)
\(\Leftrightarrow\)\(27x^3+27x^2=0\)
\(\Leftrightarrow\)\(27x^2\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy....