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a) \(\left(x+2\right)^2-9=0\)
\(=>\left(x+2\right)^2-3^2=0\\ =>\left(x+2-3\right).\left(x+2+3\right)=0\)
\(=>\left(x-1\right).\left(x+5\right)=0\)
\(=>\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}=>\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy x= 1 hoặc x= -5
b) \(x^2-2x+1=25\)
\(=>x^2-2.x.x+1^2=25\)
\(=>\left(x-1\right)^2-25=0\\ =>\left(x-1\right)^2-5^2=0\)
\(=>\left(x-1-5\right).\left(x-1+5\right)=0\)
\(=>\left(x-6\right).\left(x+4\right)=0=>\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
Vậy x= 6 hoặc x= -4
c) \(4x\left(x-1\right)-\left(2x+5\right)\left(2x-5\right)=1\)
\(=>4x\left(x-1\right)-\left[\left(2x\right)^2-5^2\right]=1\)
\(=>4x\left(x-1\right)-4x^2+25-1=0\)
\(=>4x\left(x-1\right)-4x^2+24=0\)
\(=>4x\left(x-1\right)-\left(4x^2-24\right)=0\\ =>4x\left(x-1\right)-4\left(x^2-6\right)=0\)
..................... tắc ròi -.-"
d) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+3\right)=15\)
\(=>x^3+27-x^3-3x=15\)
\(=>27-3x-15=0=>12-3x=0=>3\left(4-x\right)=0\)
Vì \(3>0=>4-x=0=>x=4\)
Vậy x= 4
e) \(3\left(x+2\right)^2+\left(2x+1\right)^2-7\left(x+3\right)\left(x-3\right)=28\)
\(=>3\left(x^2+2.x.2+2^2\right)+4x^2+4x+1-7\left(x^2-9\right)=28\)
\(=>3\left(x^2+4x+4\right)+4x^2+4x+1-7x^2+63=28\)
\(=>3x^2+12x+12+4x^2+4x+1-7x^2+63=28\)
\(=>16x+75=28=>16x=-47=>x=\frac{-47}{16}\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt :>'-'
Mình giải từ cuối lên , mình giải dần -)
n, <=> x(2x-1)-3(2x-1)=0
<=> (x-3)(2x-1)=0
<=> x= 3 hoặc x= 1/2
m, <=> (x+2)(x2-3x+5)-x2(x+2)=0
<=> (x+2)(x2-3x+5-x2)=0
<=> (x+2)(5-3x)=0
=> x= -2 hoặc5/3
\(a,\) \(3a^3.\left(x^2-1\right)^4:3a^3.\left(x^2-1\right)^3=15\)
\(\frac{3a^3.\left(x^2-1\right)^4}{3a^3.\left(x^2-1\right)^3}=15\\ x^2-1=15\\ x^2=16\\ x=4\)
\(b,\) \(x^3.\left(2x-1\right)^{m+2}:x^3.\left(2x-1\right)^{m-1}=3^5:3^2\\ \frac{x^3.\left(2x-1\right)^{m+2}}{x^3\left(2x-1\right)^{m-1}}=3^{5-2}\\ \left(2x-1\right)^3=3^3\)
\(2x-1=3\\ 2x=4\\ x=2\)
\(a.x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-5^2\right)-\left(x^3+2^3\right)=3\)
\(\Leftrightarrow x^3-25x-x^3-8=3\)
\(\Leftrightarrow x^3-x^3-25x=8+3\)
\(\Leftrightarrow x=\frac{11}{-25}\)
Vậy x có nghiệm là \(\frac{-11}{25}.\)
\(\)
Cho phép mình làm câu b trước :v
b) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
\(\Rightarrow x^6-3x^4+3x^2-1-\left(x^6-1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6+1=0\)
\(\Leftrightarrow-3x^2\cdot\left(x^2-1\right)=0\)
\(\Leftrightarrow x^2\cdot\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2=0\\x^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1;x=-1\end{matrix}\right.\)
Vậy \(x_1=-1;x_2=0;x=1\)
Xử câu a nào :D
a) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Rightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow0+3x^2+26x+28=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x+26=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x=-26\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
Vậy \(x_1=0;x_2=-\dfrac{26}{3}\)