mấy cái sau x là mũ nhé

a, \(P\left(x\right)=15-4x^3+3x^2+2x-x^3-10=-5x^3+3x^2+2x+5\)

\(Q\left(x\right)=5+4x^3+6x^2-5x-9x^3+7x=-5x^3+6x^2+2x+5\)

b, \(P\left(x\right)+Q\left(x\right)=-5x^3+3x^2+2x+5-5x^3+6x^2+2x+5\)

\(=-10x^3+9x^2+4x+10\)Thay x = 1/2 vào ta được : 

\(=-\frac{10.1}{8}+\frac{9.1}{4}+\frac{4.1}{2}+10=-\frac{5}{4}+\frac{9}{4}+2+10=1+2+10=13\)

c, \(P\left(x\right)-Q\left(x\right)=-5x^3+3x^2+2x+5+5x^3-6x^2-2x-5=6\)

\(\Leftrightarrow-3x^2=6\Leftrightarrow x^2=-2\)vô lí vì \(x^2\ge0;-2< 0\)

a, Ta có : \(5A=3B=15C\Rightarrow\frac{5A}{15}=\frac{3B}{15}=\frac{15C}{15}\Rightarrow\frac{A}{3}=\frac{B}{5}=C\)

và \(A+B+C=180^0\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{A}{4}=\frac{B}{5}=C=\frac{A+B+C}{4+5+1}=\frac{180}{10}=18\Rightarrow A=72^0;B=90^0;C=18^0\)

b, Do AD là tia phân giác ^A => \(\widehat{BAD}=\frac{1}{2}\widehat{A}=\frac{72}{2}=36^0\)

Lại có : \(\widehat{BAD}+\widehat{ADB}+\widehat{ABD}=180^0\)( tổng số đo 3 góc trong tam giác )

\(\Rightarrow\widehat{ADB}=180^0-\widehat{BAD}-\widehat{ABD}=180^0-90^0-36^0=54^0\)

nhầm r theo tính chất dãy tỉ số bằng nhau thì phải là A/3

( x - 1 )^x+2 = ( x - 1 )^x+6

=> ( x - 1 )^x+6 -(x-1)^x+2 = 0

=> ( x - 1 )^x+2 . [ ( x - 1 )⁴ - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^4-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x-1∈\left\{1;-1\right\}\end{cases}}\)

Từ x - 1 ∈ { 1 ; -1 }

\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

ta có : 

\(\frac{9a^2+4b^2}{9a^2-4b^2}=\frac{9b^2k^2+4b^2}{9b^2k^2-4b^2}=\frac{b^2\left(9k^2+4\right)}{b^2\left(9k^2-4\right)}=\frac{9k^2+4}{9k^2-4}\)

\(\frac{9c^2+4d^2}{9c^2-4d^2}=\frac{9d^2k^2+4d^2}{9d^2k^2-4d^2}=\frac{d^2\left(9k^2+4\right)}{d^2\left(9k^2-4\right)}=\frac{9k^2+4}{9k^2-4}\)

=> đpcm

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{b}=k\Rightarrow a=bk\Rightarrow a^2=b^2k^2\\\frac{c}{d}=k\Rightarrow c=dk\Rightarrow c^2=d^2k^2\end{cases}}\)

đặt VT = \(\frac{9a^2+4b^2}{9a^2-4b^2}=\frac{9b^2k^2+4b^2}{9b^2k^2-4b^2}\) \(=\frac{b^2\left(9k^2+4\right)}{b^2\left(9k^2-4\right)}=\frac{9k^2+4}{9k^2-4}\)

đặt VP = \(\frac{9c^2+4d^2}{9c^2-4d^2}=\frac{9d^2k^2+4d^2}{9d^2k^2-4d^2}\) \(=\frac{d^2\left(9k^2+4\right)}{d^2\left(9k^2-4\right)}=\frac{9k^2+4}{9k^2-4}\)

=> VT = VP

vậy 2 bt trên = nhau

ta có 5^-3/2.-2^2/5 =-2/625

=> x=1

Bài làm

Ta có: ˆxOy=ˆxOm+ˆyOn+ˆmOz+ˆzOn

Mà ˆxOm=ˆyOn=2ˆxOm

Oz là tia phân giác của ˆmOn

=> ˆmOz=ˆzOn=2ˆmOz

=> ˆxOy=2ˆxOm+2ˆmOz

Hay 1800=2ˆxOm+2ˆmOz

=> 1800=2(ˆxOm+ˆmOz)

=> ˆxOm+ˆmOz=1800:2

=> ˆxOm+ˆmOz=900xOm^+mOz^=90

Hay ˆxOz=900

=> Oz⊥xy

Vậy Oz⊥xy( đpcm )

HT NHÉ BẠN

\(M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}.\)

\(3M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+....+\frac{100}{3^{99}}\)

\(3M-M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}-\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}\)

\(2M=1+\left(\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+...+\left(\frac{100}{3^{99}}-\frac{100}{3^{99}}\right)-\frac{100}{3^{100}}\)

\(2M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt :

 \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\)

\(3N=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{98}}\)

\(3N-N=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(2N=1-\frac{1}{3^{99}}\)

\(N=\frac{1-\frac{1}{3^{99}}}{2}\)

Thay vào ta có :

\(2M=1+\frac{1-\frac{1}{3^{99}}}{2}-\frac{100}{3^{100}}\)

\(2M=1+\frac{1}{2}-\frac{1}{2\times3^{100}}-\frac{100}{3^{100}}< 1+\frac{1}{2}=\frac{3}{2}\)

Ta có : \(\frac{3}{4}< \frac{3}{2}\)

\(\Rightarrow\)\(2M=1+\frac{1}{2}-\frac{1}{2\times3^{100}}-\frac{100}{3^{100}}< \frac{3}{4}\)

\(\Rightarrow\)\(M< \frac{3}{4}\)

* Sai thì xin lỗi ạ ! *