\(M=x^4-4x+7=\left(x^2-4x+4\right)+3=\left(x-2\right)^2+3\ge3\)

\(minM=3\Leftrightarrow x=2\)

\(P=x^2-6x+y^2-2y+12=\left(x^2-6x+9\right)+\left(y^2-2y+1\right)+2=\left(x-3\right)^2+\left(y-1\right)^2+2\ge2\)

\(minP=2\Leftrightarrow\)\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

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

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

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: \(72\cdot68=4896\)

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: \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)

Tick cho mình nhé

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

a) (a+b)^3 +15a

b) -(b-2)^3

ko đăng những câu hỏi ko liên quan tới toán