\(\left(x+1\right)^2=x^2+2\cdot x\cdot1+1^2=x^2+2x+1=VP\left(đpcm\right)\)

\(P\left(x\right)=x^2+2x+4\)

\(\Delta=b^2-4ac=2^2-4\cdot1\cdot4=4-16=-12\)

\(\Delta< 0\)=> Đa thức vô nghiệm ( đpcm ) 

\(\left(x+1\right)^2=\left(x+1\right)\left(x+1\right)=x^2+x+x+1=x^2+2x+1\)

=>  \(x^2+2x+1=x^2+2x+1\left(\text{đ}pcm\right)\)

Ta có : \(P\left(x\right)=x^2+2x+4=0\)

\(\hept{\begin{cases}x^2\ge0\\2x\ge0\\4>0\end{cases}\Rightarrow vonghiem}\)

\(A=\frac{9}{10}\times\frac{10}{11}\times\frac{11}{12}\times...\times\frac{99}{100}\)

\(A=\frac{9\times10\times11\times...99}{10\times11\times12\times...\times100}\)

\(\Rightarrow A=\frac{9}{100}\)

Ta kéo Ax cắt CB tại H ( mình quyên đặt ...hihi) 

Vì Ax // Cy mà \(\widehat{C}\)= 95⁰ \(\Rightarrow\)\(\widehat{AHB}\) = 95⁰, \(\widehat{B}\)= 30⁰  \(\Rightarrow\)\(\widehat{HAB}\) = 180⁰ - 95⁰ - 30⁰ = 55

\(\Rightarrow\)\(\widehat{A}\)= 180⁰ - 55⁰( kề bù)  =  125⁰  

Vậy: \(\widehat{A}\)= 125⁰

a) M = x² + 4y² + 4xy = x² + 2xy + 2xy + 4y² = x(x + 2y) + 2y(x + 2y) = (x + 2y)² \(\ge\)0 \(\forall\)x;y

b) x = 4y, ta có: M = 9 

<=> (4y + 2y)² = 9

<=> 36y² = 9

<=> y² = 1/4

<=> \(\orbr{\begin{cases}y=\frac{1}{2}\\y=-\frac{1}{2}\end{cases}}\)

Với y = 1/2 => x = 4.1/2 = 2

y = -1/2 => x = 4. (-1/2) = -2

\(M=x^2+4y^2+4xy\)

\(\Rightarrow M=x^2+\left(2y\right)^2+2xy+2xy\)

\(\Rightarrow M=x^2+\left(2y\right)^2+\left(2xy\right)^2\ge0\forall x;y\)

\(\Rightarrow M\ge0\forall x;y\)

A - B = (x³y - 2xy + 5xy² - 3x²y² - 1) - (-2x³y - xy² + 2x²y² - xy + 8)

        = x³y - 2xy + 5xy² - 3x²y² - 1 + 2x³y + xy² - 2x²y² + xy - 8

        = (x³y + 2x³y) + (-2xy + xy) + (5xy² + xy²) + (-3x²y² - 2x²y²) + (-1 - 8)

        = 3x³y  - xy + 6xy² - 5x²y² - 9

a-b=(x^3y-2xy+5xy^2-3x^2y^2-1)-(-2x^3y-xy^2+2x^2y^2-xy+8)

        =x^3y-2xy+5xy^2-3x^2y^2-1+2x^3y+xy^2-2x^2y^2+xy-8

       =x^3y+(-2xy+xy)+(5xy^2+xy^2)+(-3x^2y^2-2x^2y^2)+(-1-8)

       =x^3y-1xy+6xy^2-5x^2y^2-9

\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{99}-1\right)\left(\frac{1}{100}-1\right)\)

\(A=\left(-\frac{9}{10}\right)\cdot\left(-\frac{10}{11}\right)\cdot\left(-\frac{11}{12}\right)\cdot....\cdot\left(-\frac{98}{99}\right)\left(-\frac{99}{100}\right)\)

\(A=\frac{\left(-9\right)\left(-10\right)\left(-11\right)...\left(-98\right)\left(-99\right)}{10\cdot11\cdot12\cdot....\cdot99\cdot100}\)

\(A=\frac{9\cdot10\cdot11\cdot....\cdot98\cdot99}{10\cdot11\cdot12\cdot...\cdot99\cdot100}=\frac{9}{100}\)

Đáp án là \(-\frac{9}{100}\)nhá