số 5 nha bạn

đề ra là 3 chữ số nhé bn, vs lại trình bày ra giùm mik vs. cám ơn

\(x^3+3x^2+3x+9=0\)

\(\Leftrightarrow x^2\left(x+3\right)+3\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\x^2+3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-3\\x^2=-3\Rightarrow ktm\end{cases}}\)

Vậy x=-3

ktm là gì vậy bạn

\(a,A=9x^2+5-6x=9x^2-6x+1+4\)

\(=\left(3x-1\right)^2+4\)

Vì: \(\left(3x-1\right)^2+4\ge4\forall x\)

\(\Rightarrow\)GTNN của A là 4 tại \(\left(3x-1\right)^2=0\Rightarrow x=\frac{1}{3}\)

b,\(B=1+x^2-x=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì: \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

\(\Rightarrow\)GTNN của B là 3/4 tại \(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)

Các phần cn lại lm tg tự nha bn

a)  \(13y\left(y-8\right)-2y+16=0\)

\(\Leftrightarrow\)\(13y\left(y-8\right)-2\left(y-8\right)=0\)

\(\Leftrightarrow\)\(\left(y-8\right)\left(13y-2\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}y-8=0\\13y-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}y=8\\y=\frac{2}{13}\end{cases}}\)

Vậy...

b)  \(-10x\left(y+2\right)-y-2=0\)

\(\Leftrightarrow\)\(\left(y+2\right)\left(-10x-1\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}y+2=0\\-10x-1=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}y=-2\\x=-\frac{1}{10}\end{cases}}\)

Vậy...

\(a,13y\left(y-8\right)-2y+16=0\)

\(\Rightarrow13y\left(y-8\right)-2\left(y-8\right)=0\)

\(\Rightarrow\left(13y-2\right)\left(y-8\right)=0\)

\(\Rightarrow\hept{\begin{cases}y=\frac{2}{13}\\y=8\end{cases}}\)

\(b,-10x\left(y+2\right)-y-2=0\)

\(-10x\left(y+2\right)-\left(y+2\right)=0\)

\(\Rightarrow\left(y+2\right)\left(-10x-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}y=-2\\x=\frac{1}{-10}\end{cases}}\)

a)  \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)

\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)\left(ac+bc+c^2\right)-a^3-b^3-c^3\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)

b)  \(VT=a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)

a,6x-3-5x+15+18x-24=24

19x-12=24

19x=36

x=36/19

c,10x-6x²+6x²-10x+21=3

0x=-18

không có x

d,3x²+3x-2x²-4x=-1-x

x²-x=-1-x

x²-x+x=-1

x²=-1

không có x thỏa mãn

b,2x²+3x²-3=5x²+5x

5x²-5x²-5x=3

-5x=3

x=\(\frac{-3}{5}\)

\(x+y=1\)

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

\(\Leftrightarrow\)\(x^2+y^2=1-2xy\)

\(x+y=1\)

\(\Leftrightarrow\)\(\left(x+y\right)^3=1\)

\(\Leftrightarrow\)\(x^3+y^3=1-3xy\)

\(H=1-3xy+3xy\left(1-2xy\right)+6x^2y^2\left(xy+y\right)\)

\(=1-6x^2y^2+6x^2y^2\left(xy+y\right)\)

\(=1-6x^2y^2\left(1-xy-y\right)\)

\(=1-6x^2y^2\left(x+y-xy-y\right)\)

\(=1-6x^2y^2\left(x-xy\right)\)

\(=1-6x^3y^2\left(1-y\right)\)

\(=1-6x^3y^2\left(x+y-y\right)\)

\(=1-6x^4y^2\)

mới ra đc đến đây