cần bốc ít nhất 10 chiếc đũa ( nếu xui xẻo )

cái này ko phải toán lớp 8

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=x^3+y^3+z^3+2xy+2xz+2yz-x^3-y^3-z^3\)

\(=2xy+2xz+2yz\)

\(=2\left(xy+xz+yz\right)\)

Đc chưa ?

Phương Đỗ Sai rùi bạn.

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)

\(=\left(x+y\right)^3+3\left(x+y+z\right)\left(x+y\right)z+z^3-x^3-y^3-z^3\)

\(=x^3+y^3+3xy\left(x+y\right)+3\left(x+y+z\right)\left(x+y\right)z+z^3-x^3-y^3-z^3\)

\(=3xy\left(x+y\right)+3\left(x+y+z\right)\left(x+y\right)z\)

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

Bạn xem lại đầu bài đi

Cho bc=16 r sao phần a lại tính bc

Ta có:

\(\left(x-1\right)\left(x+2\right)\left(x-3\right)\left(x+4\right)=144\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(x^2-x-12\right)=144\)

Đặt \(x^2-x-7=m\left(1\right)\),Ta có:

\(\Leftrightarrow\left(m+5\right)\left(m-5\right)=144\)

\(\Leftrightarrow m^2=169\Rightarrow m=13\)

Thay \(\left(1\right)=13\)

\(\Rightarrow x^2-x-7=13\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Rightarrow x=5;x=-4\)

Chon tam giác ABC 

cho mik đúng ik

a

ĐKXĐ:\(x\ne1;x\ne-1\)

\(P=\frac{x^2}{x-1}-\frac{2x^2}{x^2-1}+\frac{7}{x+1}\)

\(=\frac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x^2}{\left(x-1\right)\left(x+1\right)}+\frac{7\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^3+x^2-2x^2+7x-7}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^3-x^2+7x-7}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x-1\right)\left(x^2+7\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+7}{x+1}\)

b)

\(\left|P\right|=4\Rightarrow\orbr{\begin{cases}P=4\\P=-4\end{cases}}\)

\(\Rightarrow\frac{x^2+7}{x+1}=4\left(h\right)\frac{x^2+7}{x+1}=-4\)

Đưa về tam thức bậc 2 giải nốt

xét tam giác DBA và tam giác CDB có:

\(\dfrac{BA}{BD}=\dfrac{BD}{CD}\left(=\dfrac{1}{2}\right);\widehat{ABD}=\widehat{BDC}\)  (so le trong)

=> △DBA  ∼ △CDB (c-g-c)

=> \(\widehat{ADB}=\widehat{C}=40^0\)

a

Ta có  \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) ( đúng )

\(\Rightarrow x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=\frac{3^2}{3}=3\)

Dấu "=" xảy ra tại a=b=c=1

b

\(P=\frac{x}{\left(x+10\right)^2}\)

Đặt \(y=\frac{1}{x+10}\Rightarrow x=\frac{1}{y}-10\)

\(\Rightarrow P=\left(\frac{1}{y}-10\right)\cdot y^2=-10y^2+y\)

\(=-10\left(y^2-2\cdot y\cdot\frac{1}{20}\cdot y+\frac{1}{400}\right)+\frac{1}{40}\)

\(=-10\left(y-\frac{1}{2}\right)^2+\frac{1}{40}\le\frac{1}{40}\)

Dấu "=" xảy ra tại \(y=\frac{1}{2}\Leftrightarrow x=10\)

Vậy...............................

Điều kiện xác định x khác 1

\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)

\(\Leftrightarrow\frac{1.\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow x^2+x+1-3x^2=2x^2-2x\)

\(\Leftrightarrow x^2-3x^2-2x^2+x+2x+1=0\)

\(\Leftrightarrow-4x^2+2x+1=0\)

\(\Leftrightarrow\left(-2x-1\right)^2=0\)

\(\Leftrightarrow-2x-1=0\)

\(\Rightarrow x=-0,5\)(thỏa mãn)

ok cám ơn bạn rất nhiều!