\(C=50^2-49^2+48^2-47^2+...+2^2-1^2\)

\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=50+49+48+47+...+2+1\)

\(=\frac{50\left(50+1\right)}{2}=1275\)

chỉnh đề câu a) c/m: AH² = BH.CH

a) Xét tam giác AHB và tam giác CHA có:

    góc AHB = góc CHA = 90⁰

   góc BAH = góc ACH (cùng phụ góc HAC)

suy ra:  tam giác AHB ~ tam giác CHA  (g.g)

=>    \(\frac{AH}{CH}=\frac{HB}{AH}\)    =>  AH² = HB.CH

b) CH = BC - BH = 13 - 4 = 9

Áp dụng câu a) ta có:  \(AH^2=HB.CH\)

                         \(\Rightarrow\)\(AH=\sqrt{HB.CH}=\sqrt{4.9}=6\)

CM hộ mik với :v

\(81x^3-27\)

\(=\left(\sqrt[3]{81}.x\right)^3-3^3\)

\(=\left(\sqrt[3]{81}.x-3\right)\left[\left(\sqrt[3]{81}.x\right)^2+\sqrt[3]{81}.x.3+9\right]\)

Tham khảo nhé~

a)4-x^2-4x=4-x^2-2x-2x=-2x-4-x*(x+2)=-2*(x+2)-x*(x+2)=(-2-x)*(x+2)=-(x+2)^2

b)81x^3-27=27*3*x^3-27=27(3*x^3-1)

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

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

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

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

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

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

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

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

\(=0\)    do a+b+c = 0

Ta có: \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)

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

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

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

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

\(=-abc.\left(c-a\right)-abc\left(b-c\right)-abc\left(a-b\right)\)

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

\(=-abc.0\)

\(=0\)

Vậy \(\left(a-b\right).c^3+\left(c-a\right)b^3+\left(b-c\right)a^3=0\)

Tham khảo nhé~

\(P=n^3-n^2+n-1\)

\(=n^2\left(n-1\right)+\left(n-1\right)\)

\(=\left(n-1\right)\left(n^2+1\right)\)

Đế P là số nguyên tố thì:  \(\orbr{\begin{cases}n-1=1\\n^2+1=1\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}n=2\left(TM\right)\\n=0\left(L\right)\end{cases}}\)

Vậy n= 2

"."

Có mk nè !

.

hình đâu bạn 

Không có hình

a) \(5x^2-12xy+9y^2-4x+4=\left(4x^2-12xy+9y^2\right)+x^2-4x+4=\left(2x-3y\right)^2+\left(x-2\right)^2\ge0\)
b) \(-x^2-2y^2+12x-4y+7=-\left(x^2-12x+36\right)-2\left(y^2+2y+1\right)+45=-\left(x-6\right)^2-2\left(y+1\right)^2+45\le45\)

c)\(4y^2+10x^2+12xy+6x+7=\left(4y^2+12xy+9x^2\right)+x^2+6x+9-2=\left(2y+3x\right)^2+\left(x+3\right)^2-2\ge-2\)

d) \(3-10x^2-4xy-4y^2=3-\left(4y^2+4xy+x^2\right)-9x^2=-\left(2y+x\right)^2-9x^2+3\le3\)

e)\(x^2-5x+y^2-xy-4y+16=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\frac{1}{2}\left(x^2-10x+25\right)+\frac{1}{2}\left(y^2-8y+16\right)-\frac{9}{2}=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-5\right)^2+\frac{1}{2}\left(y-4\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)Phần e) mới nghĩ đk v, tui biết đáp án sao do k xảy ra dấu bằng