\(f\left(10\right)=100a+10b+c\)

\(=10\left(7a+b\right)+30a+c\)

\(=30a+c\)

\(f\left(-3\right)=9a-3b+c\)

\(=7a+b+2a-4b+c\)

\(=2a+28a+c\)

\(=30a+c\)

\(f\left(-3\right).f\left(10\right)=\left(30a+c\right)\left(30a+c\right)\)

a) tamgiac ABC cân tại A nên: AB=AC=6(cm)

và gócB=gócC

b) cạnh BC không thể bằng cạnh CD được. Em xem lại đề nhé

a)

$3x^2-2x-1=0\iff \left(3x^2-3x\right)+\left(x-1\right)=0$

$\iff \left(x-1\right)\left(3x+1\right)=0\iff [\begin{array}{c}x=1\\ x=\frac{-1}{3}\end{array}$

b)

$\iff \left(x^4-x^2\right)+\left(4x^2+4\right)=0$

$\iff x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-1\right)\left(x^2-4\right)=0$

$\iff [\begin{array}{c}{x}^2-1=0\\ x^2-4=0\end{array}$ $\iff [\begin{array}{c}x=-1\\ x=1\\ x=-2\end{array}$$\begin{array}{c}x=2\end{array}$

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

\(\Rightarrow3x^2-x+3x-1=0\)

\(\Rightarrow x\left(3x-1\right)+\left(3x-1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)

Vậy nghiệm của đa thức là x=-1 và x=1/3

\(\)

Thay x = -2 và B = 4 ta có : 

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

\(< =>2m=4-4-1\)

\(< =>2m=-1< =>m=-\frac{1}{2}\)

điều kiện xác định: \(\hept{\begin{cases}x\ne0\\x\ne9\end{cases}}\)

PT <=> \(\frac{180\left(x+9\right)+180x+x\left(x+9\right)}{2x\left(x+9\right)}=5\)

\(\Leftrightarrow180x+1620+180x+x^2+9x=10x^2+90x\)

\(\Leftrightarrow9x^2-279x-1620=0\)

\(\Leftrightarrow x^2-31x-180=0\)

\(\Leftrightarrow\left(x-36\right)\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=36\\x=-5\end{cases}}\)

Vậy ...

\(\frac{90}{x}+\frac{90}{x+9}+\frac{1}{2}=5\)\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-9\end{cases}}\)

\(\Leftrightarrow\frac{180\left(x+9\right)}{2x\left(x+9\right)}+\frac{180x}{2x\left(x+9\right)}+\frac{x\left(x+9\right)}{2x\left(x+9\right)}=\frac{10x\left(x+9\right)}{2x\left(x+9\right)}\)

\(\Rightarrow180x+1620+180x+x^2+9x=10x^2+90x\)

\(\Leftrightarrow180x+1620+180x+x^2+9x-10x^2-90x=0\)

\(\Leftrightarrow-9x^2+279x+1620=0\)

\(\Leftrightarrow\left(x-36\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-36=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=36\\x=-5\end{cases}}}\)

\(a)\)

\(\text{Ta có}:\)

\(\Delta ABC\)\(\text{vuông tại}\)\(A\)

\(\rightarrow BC^2=AB^2+AC^2\)

\(\rightarrow AC^2=BC^2-AB^2\)

\(\rightarrow AC^2=15^2-9^2\)

\(\rightarrow AC^2=144\)

\(\rightarrow AC=12\)

\(\rightarrow AB< AC< BC\)

\(\rightarrow\widehat{C}< \widehat{B}< \widehat{A}\)

\(\text{Ta có:}\)

\(AB\perp AC\rightarrow\widehat{BAC}=\widehat{EAC}\)

\(\rightarrow AB=AE\rightarrow A\)\(\text{là trung điểm}\)\(BE\)

\(b)\)

\(\text{Theo phần a), ta có:}\)\(AB=AE\rightarrow A\text{ }\)\(\text{là trung điểm}\)\(BE\)
\(\rightarrow CA\)\(\text{là trung tuyến}\)\(\Delta CBE\)

\(\text{Mà}\)\(BH\)\(\text{là trung tuyến}\)\(\Delta BCE\)\(,\)\(BH\text{∩}\text{ }CA=M\)

\(\rightarrow M\text{ }\)\(\text{là trọng tâm}\)\(\Delta BCE\)

\(\rightarrow CM=\frac{2}{3}CA\)

\(\rightarrow CM=8\)

\(c)\)

\(\text{Theo phần a)}\)\(\rightarrow\widehat{ECA}=\widehat{ACB}\)

                         \(\rightarrow\widehat{CEA}=\widehat{CBA}\)

\(\text{Do}\)\(AK//CE\rightarrow\widehat{KAB}=\widehat{AEC}=\widehat{CBA}=\widehat{KBA}\rightarrow KB=KA\)

         \(\widehat{KAC}=\widehat{ECA}=\widehat{ACB}=\widehat{ACK}\rightarrow KA=KC\)

         \(\rightarrow KB=KC\rightarrow K\)\(\text{là trung điểm}\)\(BC\)

\(\text{Mà}\)\(M\)\(\text{là trọng tâm}\)\(\Delta CBE\rightarrow E,MK\)\(\text{thẳng hàng}\)

P = x³ - 2x²y - 3x² - 2xy + 4y² + 3x - 5

= (x³ - 2x²y - 3x²) - (2xy - 4y² - 6y) + (3x - 6y - 9) + 4

= x²(x - 2y - 3) - 2y(x - 2y - 3) + 3(x - 2y-  3) + 4

= (x - 2y - 3)(x² - 2y + 3) + 4 

= 4 (Vì x - 2y - 3 = 0)

adasd

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