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TP
2
M
25 tháng 10 2017
\(\left(6x^3-7x^2-x+2\right):\left(2x+1\right).\)
\(=\left(x+\frac{1}{2}\right).\left(x-1\right).\left(x-\frac{2}{3}\right):\left(x+\frac{1}{2}\right)\)
\(=\left(x-1\right).\left(x-\frac{2}{3}\right)\)
20 tháng 8 2021
1: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-3x^2+27x-27-x^3+27+9x^2+18x+9=15\)
\(\Leftrightarrow45x=6\)
hay \(x=\dfrac{2}{15}\)
2: Ta có: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x^3-25x-x^3-8=3\)
\(\Leftrightarrow-25x=11\)
hay \(x=-\dfrac{11}{25}\)
3: Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x-5\right)\left(x+5\right)=264\)
\(\Leftrightarrow x^3+64-x^3+25x=264\)
\(\Leftrightarrow25x=200\)
hay x=8
4: Ta có: \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+8+6x^2-24=60\)
\(\Leftrightarrow12x=84\)
hay x=7
20 tháng 8 2021
6: Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=64\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8=64\)
\(\Leftrightarrow12x^2=48\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
7: Ta có: \(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x=-10\)
hay x=1
8: Ta có: \(\left(4x+1\right)^2-\left(2x+3\right)^2+5\left(x+2\right)^2+3\left(x-2\right)\left(x+2\right)=500\)
\(\Leftrightarrow16x^2+8x+1-4x^2-12x-9+5x^2+20x+20+3x^2-12=500\)
\(\Leftrightarrow20x^2+16x-500=0\)
\(\text{Δ}=16^2-4\cdot20\cdot\left(-500\right)=40256\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-16-8\sqrt{629}}{40}=\dfrac{-2-\sqrt{629}}{5}\\x_2=\dfrac{-16+8\sqrt{629}}{40}=\dfrac{-2+\sqrt{629}}{5}\end{matrix}\right.\)
9: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
17 tháng 8 2021
Bài 3:
1: \(35^2=1225\)
2: \(25^2=625\)
3: \(75^2=5625\)
4: \(95^2=9025\)
5: \(101\cdot99=9999\)
6: \(36\cdot44=1584\)
7: \(72\cdot68=4896\)
17 tháng 8 2021
1: \(x^2-2x+1=\left(x-1\right)^2\)
2: \(4x^2-4x+1=\left(2x-1\right)^2\)
3: \(16x^2+8x+1=\left(4x+1\right)^2\)
4: \(9x^2+12x+4=\left(3x+2\right)^2\)
5: \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
24 tháng 8 2021
a) \(99^3=\left(100-1\right)^3=100^3-3.100^2+3.100-1=1000000-30000+300-1=970299\)b) \(91^3+3.91^2.9+3.91.9^2+9^3=\left(91+9\right)^3=100^3=1000000\)
c) \(1001^3=\left(1000+1\right)^3=1000^3+3.1000^2+3.1000+1=1003003001\)d) \(102^3-6.102^2+24.102-8=\left(102-2\right)^3+12.102=100^3+1224=1001224\)
nhỏ quá bn ạ
\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\)
\(=\left(x^2+4x+4-3\right)^2-12\left(x+2\right)^2+2093\)
\(=\left[\left(x+2\right)^2-3\right]^2-12\left(x+2\right)^2+2093\)
\(=\left(x+2\right)^4-6\left(x+2\right)^2+9-12\left(x+2\right)^2+2093\)
\(=\left(x+2\right)^4-18\left(x+2\right)^2+2102\)
\(=\left(x+2\right)^4-18\left(x+2\right)^2+81+2021\)
\(=\left[\left(x+2\right)^4-18\left(x+2\right)^2+81\right]+2021\)
\(=\left[\left(x+2\right)^2-9\right]^2+2021\)
\(=\left[\left(x+2-3\right)\left(x+2+3\right)\right]^2+2021\)
\(=\left[\left(x-1\right)\left(x+5\right)\right]^2+2021\)
Vì \(\left[\left(x-1\right)\left(x+5\right)\right]^2\ge0\forall x\)
\(\Rightarrow\left[\left(x-1\right)\left(x+5\right)\right]^2+2021\ge2021\)\(\forall x\)
hay \(P\ge2021\)
Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy \(minP=2021\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)