thanh niên ko chịu làm bài tập mà lên đây hỏi à :))

a) Xét ΔMNP có: \(NP^2=5^2=25\)

\(MP^2+MN^2=3^2+4^2=25\)

\(\Rightarrow NP^2=MP^2+MN^2\)

\(\Rightarrow\Delta MNP\) vuông tại M (theo định lí Pi-ta-go đảo)

b) Chu vi \(\Delta MNP\) bằng: 

\(MP+MN+NP=3+4+5=12\left(cm\right)\)

Diện tích \(\Delta MNP\) bằng: 

\(\dfrac{MP\cdot MN}{2}=\dfrac{3\cdot4}{2}=6\left(cm\right)\)

Thì bạn chỉ cần trả lời đúng,trình bày đẹp,dễ hiểu thôi và chăm chỉ trả lời

Mấy cái bạn nói mình đều làm rồi có được đâu

\(x^4+3x^3-x-3=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^3-1=0\end{matrix}\right.\)            \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Vậy \(x\in\left\{-3;1\right\}\)

\(x^5+x^4+x+1=0\)

\(\Leftrightarrow x^4\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^4+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^4+1=0\end{matrix}\right.\)             \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4=-1\end{matrix}\right.\)

\(\Rightarrow x=-1\) (do \(x^4\ge0\forall x\))

Vậy x = -1

\(x^5+x^4+x+1=0\)

\(\Leftrightarrow\left(x^5+x^4\right)+\left(x+1\right)=0\)

\(\Leftrightarrow x^4\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^4+1\right)\left(x+1\right)=0\) (Mà: \(x^4+1\ge1>0\forall x\))

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

\(a)4x^2-6x\)

\(=2x\left(2x-3\right)\)

\(b)9x^4y^3+3x^2y^4\)

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

\(c)x^3-2x^2+5x\)

\(=x\left(x^2-2x+5\right)\)

\(d)3x(x-1)+5(x-1)\)

\(=\left(x-1\right)\left(3x+5\right)\)

\(e)2x^2(x+1)+4(x+1)\)

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

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

\(f)-3x-6xy+9xz\)

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

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

a) \(4x^2-6x\)

\(=2x\left(2x-3\right)\)

b) \(9x^4y^3+3x^2y^4\)

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

c) \(x^3-2x^2+5x\)

\(=x\left(x^2-2x+5\right)\)

d) \(3x\left(x-1\right)+5\left(x-1\right)\)

\(=\left(x-1\right)\left(3x+5\right)\)

e) \(2x^2\left(x+1\right)+4\left(x+1\right)\)

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

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

f) \(-3x-6xy+9xz\)

\(=-\left(3x+6xy+9xz\right)\)

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

\(xyz+xz+yz-z+xy+x-y-1\)

\(=\left(xyz+xz+yz-z\right)+\left(xy+x-y-1\right)\)

\(=z\left(xy+x-y-1\right)+\left(xy+x-y-1\right)\)

\(=\left(z+1\right)\left(xy+x-y-1\right)\)

Thay x=-9, y=-21 và x=-31 vào ta có:

\(=\left(-31+1\right)\cdot\left\{\left[-9\cdot\left(-21\right)\right]+\left(-9\right)-\left(-21\right)-1\right\}\)

\(=-31\cdot\left(180+21-1\right)\)

\(=-30\cdot200\)

\(=-6000\)

\(B=xyz+xz-yz-z+xy+x-y-1\)

\(=\left(xyz+xz-yz-z\right)+\left(xy+x-y-1\right)\)

\(=z\left(xy+x-y-1\right)+\left(xy+x-y-1\right)\)

\(=\left(xy+x-y-1\right)\left(z+1\right)\)

Với \(x=-9;y=-21;z=-31\), ta được:

\(B=\left[-9\cdot\left(-21\right)-9-\left(-21\right)-1\right]\cdot\left(-31+1\right)\)

\(=200\cdot\left(-30\right)\)

\(=-6000\)

Ta có:

\(VT=3\left(x^2+y^2+z^2\right)-\left(x-y\right)^2-\left(z-x\right)^2-\left(y-z\right)^2\)

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

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

 \(=x^2+y^2+z^2+2xy+2xz+2yz\)

\(=\left(x+y+z\right)^2=VP\) 

⇒ Đpcm

\(3\left(x^2+y^2+z^2\right)-\left(x-y\right)^2-\left(z-x\right)^2\)

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

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

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

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

Mong bạn kiểm tra lại đề bài!

\(x^4-x^2+10x-25\)

\(=x^4-\left(x^2-10x+25\right)\)

\(=\left(x^2\right)^2-\left(x^2-2\cdot5\cdot x+5^2\right)\)

\(=\left(x^2\right)^2-\left(x-5\right)^2\)

\(=\left[x^2-\left(x-5\right)\right]\left[x^2+\left(x-5\right)\right]\)

\(=\left(x^2-x+5\right)\left(x^2+x-5\right)\)

\(x^3-4x^2+8x-8\)

\(=x^3-2x^2+4x-2x^2+4x-8\)

\(=\left(x^3-2x^2+4x\right)-\left(2x^2-4x+8\right)\)

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

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

giải tiếp cách kia mà ạ