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a) (2x-3)2=3-2x
=> (3-2x)2=3-2x
=>(3-2x)(3-2x)=3-2x
=>(3-2x)(3-2x)-(3-2x)=0
=>(3-2x)(3-2x+1)=0
=>3-2x=0 hoặc 3-2x+1=0(bạn tự tính ra nha)
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ta lập bảng xét dấu, sau khi lập xong , ta xét từng trường hợp là được ( câu a)
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\(\left(x-2\right)^{2x+3}=\left(x-2\right)^{2x+1}\)
\(\Rightarrow2x+3=2x+1\)
\(\Rightarrow2x-2x=1-3\)
\(\Rightarrow0=-2\left(\text{vô lí}\right)\)
Vậy \(x=\varnothing\)
\(\left(x-2\right)^{2x+3}=\left(x-2\right)^{2x+1}\)
\(\Rightarrow\left(x-2\right)^{2x+3}-\left(x-2\right)^{2x+1}=0\)
\(\Rightarrow\left(x-2\right)^{2x+1}\left[\left(x-2\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-2\right)^{2x+1}=0\\\left(x-2\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x-2=\pm1\Rightarrow x=3\text{ }or\text{ }x=1\end{cases}}\)
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Hằng đẳng thức đó bn:
\(\left(a+b\right)\left(a^2-ab+b^2\right)\)
Thay vào thì: \(-\left(x-3\right)\left(x^2-3x+9\right)=-\left[\left(x-3\right)\left(x^2-3x+3^2\right)\right]\)
\(=-\left(x^3-27\right)=-x^3+27\)
Bài làm:
Ta có: \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)=\left(x-3\right)^3+3\left(2x+1\right)^2-\left(x^3-5x+1\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3+27=x^3-9x^2+27x-27+12x^2+12x+3-x^3+5x-1\)
\(\Leftrightarrow6x^2+41x-51=0\)
\(\Leftrightarrow6\left(x^2+\frac{41}{6}x+\frac{1681}{144}\right)-\frac{2905}{24}=0\)
\(\Leftrightarrow\left(x+\frac{41}{12}\right)^2-\frac{\left(\sqrt{2905}\right)^2}{12^2}=0\)
\(\Leftrightarrow\left(x+\frac{41}{12}-\frac{\sqrt{2905}}{12}\right)\left(x+\frac{41}{12}+\frac{\sqrt{2905}}{12}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{2905}-41}{12}\\x=\frac{-\sqrt{2905}-41}{12}\end{cases}}\)
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\(3^{2x}+3^{2x+1}=324\)
\(\Rightarrow3^{2x}+3^{2x}.3=324\)
\(\Rightarrow3^{2x}\left(1+3\right)=324\)
\(\Rightarrow3^{2x}.4=324\)
\(\Rightarrow3^{2x}=324:4=81=3^4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2=2\)
32x + 32x + 1 = 324
=> 32x + 32x.3 = 324
=> 32x.(1 + 3) = 324
=> 32x.4 = 324
=> 32x = 324 : 4
=> 32x = 81
=> 32x = 34
=> 2x = 4
=> x = 2
=> x+3=x+1
=>x-x= -3+1
=> -x= -2
=> x=2
\(\left(2x-3\right)^{x+3}=\left(2x-3\right)^{x+1}\)
=> \(\left(2x-3\right)^{x+3}-\left(2x-3\right)^{x+1}=0\)
=> \(\left(2x-3\right)^{x+1}\left[\left(2x-3\right)^2-1\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-3\right)^{x+1}=0\\\left(2x-3\right)^2-1=0\end{cases}}\)
= > \(\orbr{\begin{cases}2x-3=0\\\left(2x-3\right)^2=1\end{cases}}\)
=> 2x = 3
hoặc : 2x - 3 = 1 hoặc 2x -3 = -1
=> x = 3/2
hoặc x = 2 hoặc x = 1
Vậy ...