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a) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)
\(\Leftrightarrow x^3-27+4x-x^3=1\)
\(\Leftrightarrow4x-27=1\Leftrightarrow4x=28\Leftrightarrow x=7\)
b) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)
\(\Leftrightarrow x^3-6x^2+12x+4-x^3-8+6\left(x^2-4\right)=60\)
\(\Leftrightarrow-6x^2+12x-4+6x^2-24=60\)
\(\Leftrightarrow12x-28=60\Leftrightarrow x=\frac{22}{3}\)
a) (x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1
x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 + 2x^2 - x^3 + 4x - 2x^2 = 1
4x - 27 = 1
4x = 28
x = 7
b) (x - 2)^3 - (x - 2)(x^2 + 2x + 4) + 6(x - 2)(x + 2) = 60
x^3 - 4x^2 + 4x - 2x^2 + 8x - 8 - x(x^2 + 2x + 4) + 2(x^2 + 2x + 4) + 6x - 24 = 60
x^3 + 12x - 32 - x^2 - 2x^2 - 4x + 2x^2 + 4x + 8 = 60
12x - 24 = 60
12x = 60 + 24
12x = 84
x = 7
a) x^4 - 2x^2 + 1 = 0
=> ( x^2 - 1 )^2 = 0
=> x^2 - 1 = 0
=> x^2 = 1
=> x = 1 hoặc x = -1
a) x4-2x2+1=0
(thang Tran giải rồi nhé)
b) x4-2x2-8=0
<=> x^4 - 2x^2 +1 -9 =0
<=> (x^2 -1)^2 -9 =0
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=-3\\x^2-1=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-2\left(VN\right)\\x=+_-\sqrt{2}\end{cases}}}\)
Vậy x=+- căn 2
c) x4-4x2-60=0
\(\Leftrightarrow x^4-4x^2+4-64=0\)
\(\Leftrightarrow\left(x^2-2\right)-64=0\)
\(\Leftrightarrow\left(x^2+62\right)\left(x^2-66\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+62=0\\x^2-66=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-62\left(VN\right)\\x^2=+_-\sqrt{66}\end{cases}}}\)
Vậy x=+- căn 66
d) x6-16x2+64=0
a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow4x=7\Leftrightarrow x=1,75\)
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10.\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\)
\(\Leftrightarrow24x=27\Leftrightarrow x=1,125\)
\(12\left(x-2\right)\left(x+2\right)-3\left(2x+3\right)^2\) \(=52\)
\(12\left(x^2-4\right)-3\left(4x^2+12x+9\right)\) \(=52\)
\(12x^2-48-12x^2-36x-27\) \(=52\)
\(-36x-75=52\)
\(-36x=127\)
\(x=\frac{-127}{36}\)
\(\left(2x+1\right)^2-4\left(x-1\right)\left(x+1\right)\) \(+2x=5\)
\(4x^2+4x+1-4\left(x^2-1\right)\) \(+2x=5\)
\(4x^2+4x-1-4x^2+4+2x=5\)
\(6x+3=5\)
\(6x=2\)
\(x=3\)
\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\) \(+6\left(x-1\right)^2=15\)
\(x^3-6x^2+12x-8-\left(x-3\right)\left(x+3\right)^2\) \(+6\left(x^2-2x+1\right)=15\)
\(x^3-6x^2+12x-8-\left(x^2-9\right)\left(x+3\right)\) \(+6x^2-12x+6=15\)
\(x^3-2\) \(-\left(x^3+3x^2-9x-27\right)\)\(=15\)
\(x^3-2-x^3-3x^2+9x+27=15\)
\(-3x^2+9x+25=15\)
\(-3x^2+9x+10=0\)
\(-3\left(x^2-3x-\frac{10}{3}\right)\) \(=0\)
\(x=\frac{9+\sqrt{201}}{6}\)
các câu còn lại tương tự
a ) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-x^2+9=6\)
\(\Leftrightarrow-4x+13=6\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\frac{7}{4}\)
Vậy \(x=1\).
b ) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)
\(\Leftrightarrow-24x+37=10\)
\(\Leftrightarrow-24x=27\)
\(\Leftrightarrow x=\frac{9}{8}.\)
Mấy pài kia tương tự . :D
Câu 2 sai đề nhé
Phải là:(x-999)/99+(x-896)/101+(x-789/103)=6
\(\left(x-2\right)^3-\left(x-2x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)
\(\Rightarrow x^3-6x^2+12x-8-x+2x^2-2x-4+6\left(x^2-4\right)=60\)
\(\Rightarrow x^3-6x^2+12x-8-x+2x^2-2x-4+6x^2-24=60\)
\(\Rightarrow x^3+2x^2-7x-36=60\)
\(\Rightarrow x^3+2x^2-7x-96=0\)
Sai đề không ???
k sai đề nha