làm a,b,c thôi ạ
gấpppppppppppppppppppppppppppppppp
Câu d thôi ạ, câu a, b, c làm rồi ạ
làm phần b,c,d thôi ạ
\(b,\left(x+5\right)^2=100\)
TH1:\(\left(x+5\right)^2=10^2\)
\(\Rightarrow x+5=10\\ \Rightarrow x=10-5\\ \Rightarrow x=5\)
TH2:\(\left(x+5\right)^2=\left(-10\right)^2\)
\(\Rightarrow x+5=-10\\ \Rightarrow x=-10-5\\ \Rightarrow x=-15\)
\(c,\left(2x-4\right)^2=0\\ \Rightarrow2x-4=0\\ \Rightarrow2x=0+4\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(d,\left(x-1\right)^3=-27\\ \Rightarrow\left(x-1\right)^3=\left(-3\right)^3\\ \Rightarrow x-1=-3\\ \Rightarrow x=-3+1\\ \Rightarrow x=-2\)
b) (x+5)2=100
(x+5)2=102
x+5=10
x=10-5
x=5
c) (2x-4)2=0
2x-4=0
2x=4
x=4:2
x=\(\pm\)2
d)(x-1)3=-27
(x-1)3=-33
x-1=-3
x=-3+1
x=-2
làm hộ câu b,c,d thôi ạ
b: \(=\dfrac{x^2-x+1-3+1-x^2}{\left(x+1\right)\cdot\left(x^2-x+1\right)}=\dfrac{-x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{-1}{x^2-x+1}\)
Ý a,b,c cho mình xin kết quả để đối chiếu thôi ạ. Còn ý d làm rõ hộ mình. Cảm ơn nhiều ạ❤
a. \(f\left(x\right)_{max}=f\left(-2\right)=111\) ; \(f\left(x\right)_{min}=f\left(1\right)=-6\)
b. \(f\left(x\right)_{max}=f\left(-3\right)=7\) ; \(f\left(x\right)_{min}=f\left(0\right)=1\)
c. \(f\left(x\right)_{max}=f\left(4\right)=\dfrac{2}{3}\) ; \(f\left(x\right)_{min}\) ko tồn tại
d.
Miền xác định: \(D=\left[-2\sqrt{2};2\sqrt{2}\right]\)
\(y'=\dfrac{2\left(4-x^2\right)}{\sqrt{8-x^2}}=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
\(f\left(-2\sqrt{2}\right)=f\left(2\sqrt{2}\right)=0\)
\(f\left(-2\right)=-4\) ; \(f\left(2\right)=4\)
\(f\left(x\right)_{max}=f\left(2\right)=4\) ; \(f\left(x\right)_{min}=f\left(-2\right)=-4\)
làm nhanh giúp mình câu b và c thôi ạ
câu a thôi ạ, câu b làm rồi
`a)P(x)+Q(x)=x^5-2x^2+1`
`=>Q(x)=x^5-2x^2+1-P(x)`
`=>Q(x)=x^5-2x^2+1-x^4+3x^2-1/2+x`
`=>Q(x)=x^5-x^4+x^2+x+1/2`
______________________________________________
`b)P(x)-R(x)=x^3`
`=>R(x)=P(x)-x^3`
`=>R(x)=x^4-3x^2+1/2-x-x^3`
`=>R(x)=x^4-x^3-3x^2-x+1/2`
Ta có:
\(P\left(x\right)+Q\left(x\right)=x^5-2x^2+1\)
\(\Rightarrow Q\left(x\right)=P\left(x\right)-\left(x^5-2x^2+1\right)\)
\(=x^4-3x^2+\dfrac{1}{2}-x-x^5+2x^2-1\)
\(=-x^5+x^4-x^2-x-\dfrac{1}{2}\)
Vậy \(Q\left(x\right)=-5^2+x^4-x^2-x-\dfrac{1}{2}\)
a) <=> Q(x) = (x5 - 2x2 + 1) - P(x)
= (x5 - 2x2 + 1) - (x4 - 3x2 + 1/2 - x)
= x5 - 2x2 + 1 - x4 + 3x2 + x - 1/2
= x5 - x4 + x2 + x + 1/2
Vậy Q(x) = x5 - x4 + x2 + x + 1/2
Cho a, b, c>0; abc=1. Cmr:
\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c^3}{a\left(b+2\right)}\ge1\)
Sao em làm chỉ ra >=3 thôi ạ)):
\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b}{3}+\dfrac{c+2}{9}\ge3\sqrt[3]{\dfrac{a^3b\left(b+2\right)}{27b\left(c+2\right)}}=a\)
Tương tự: \(\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c}{3}+\dfrac{a+2}{9}\ge b\)
\(\dfrac{c^3}{a\left(b+2\right)}+\dfrac{a}{3}+\dfrac{b+2}{9}\ge c\)
Cộng vế:
\(VT+\dfrac{4\left(a+b+c\right)}{9}+\dfrac{2}{3}\ge a+b+c\)
\(\Rightarrow VT\ge\dfrac{5\left(a+b+c\right)}{9}-\dfrac{2}{3}\ge\dfrac{15}{9}-\dfrac{2}{3}=1\)
Mấy bạn làm giúp mik câu b và c thôi ạ, mik cảm ơn
Gấp ạ!!! Làm câu c thôi ạ
c: Ta có: AM//BC
AE⊥BC
Do đó:AM⊥AE
Suy ra: \(\widehat{AME}+\widehat{AEM}=90^0\)
hay \(\widehat{AME}+\widehat{BAD}=90^0\)