16,2 nhe . nho k mik ^ - ^

ta có: 30,9-28,2+25,5-22,8+....+3,9-1,2

=(30,9+25,5 +...+ 3,9)-(28,2+22,8+...+1,2)

=[(30,9+3,9)*6 : 2 ]-[(28,2+1,2)*6 : 2]

= 104,4-88,2

=16,2

16,2 bạn nhé

Câu 1: \(\frac{\pi}{2}<\alpha,\beta<\pi\)

=>\(\sin\alpha>0;\sin\beta>0;cos\alpha<0;cos\beta<0\)

\(\sin^2\alpha+cos^2\alpha=1\)

=>\(cos^2\alpha=1-\sin^2\alpha=1-\left(\frac13\right)^2=\frac89\)

mà \(cos\alpha<0\)

nên \(cos\alpha=-\frac{2\sqrt2}{3}\)

Ta có: \(\sin^2\beta+cos^2\beta=1\)

=>\(\sin^2\beta=1-\left(-\frac23\right)^2=1-\frac49=\frac59\)

mà \(\sin\beta>0\)

nên \(\sin\beta=\frac{\sqrt5}{3}\)

\(\sin\left(\alpha+\beta\right)=\sin\alpha\cdot cos\beta+cos\alpha\cdot\sin\beta\)

\(=\frac13\cdot\frac{-2}{3}+\frac{-2\sqrt2}{3}\cdot\frac{\sqrt5}{3}=\frac{-\sqrt2-2\sqrt{10}}{9}\)

Câu 2:

\(P=cos\left(a+b\right)\cdot cos\left(a-b\right)\)

\(=\frac12\cdot\left\lbrack cos\left(a+b+a-b\right)+cos\left(a+b-a+b\right)\right\rbrack=\frac12\cdot\left\lbrack cos2a+cos2b\right\rbrack\)

\(=\frac12\cdot\left\lbrack2\cdot cos^2a-1+2\cdot cos^2b-1\right\rbrack=cos^2a+cos^2b-1\)

\(=\left(\frac13\right)^2+\left(\frac14\right)^2-1=\frac19+\frac{1}{16}-1=\frac{25}{144}-1=-\frac{119}{144}\)

1.

\(P=\dfrac{3}{\sqrt{x}+3}\le\dfrac{3}{0+3}=1\)

\(\Rightarrow maxP=1\Leftrightarrow x=0\)

2. 

\(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}=2-\dfrac{3}{\sqrt{x}+1}\ge2-\dfrac{3}{0+1}=-1\)

\(\Rightarrow minA=-1\Leftrightarrow x=0\)

\(a,\Leftrightarrow\Delta'=\left(m-1\right)^2-\left(m-2\right)\ge0\\ \Leftrightarrow m^2-3m+3\ge0\\ \Leftrightarrow\left(m-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge0\left(\text{luôn đúng}\right)\)

Vậy PT có 2 nghiệm pb với mọi m

\(b,\Leftrightarrow0< x_1< x_2\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2>0\\x_1x_2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(m-1\right)>0\\m-2>0\end{matrix}\right.\Leftrightarrow m>2\\ c,\text{Thay }x=2\Leftrightarrow4-4\left(m-1\right)+m-2=0\\ \Leftrightarrow m=2\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\text{Viét: }\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=m-2\end{matrix}\right.\\ x_1^2+x_2^2=8\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=8\\ \Leftrightarrow4\left(m-1\right)^2-2\left(m-2\right)=8\\ \Leftrightarrow4m^2-10m=0\\ \Leftrightarrow m\left(2m-5\right)=0\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

 X x 0,25 + X : 0,25 = 4,59

X x 0,25 + X x 4 = 4,59

X x ( 0,25 + 4 ) = 4,59

X x 4,25 = 4,59

X = 4,59 : 4,25

X = 1,08

Bài 2 :

X : 0,1 - 9,9 = X x 0,1 - 9,9

X x 10 - 9,9 =  X x 0,1 - 9,9

X x 10 - X x 0,1 = 9,9 - 9,9

X x ( 10 - 0,1 ) = 0

X x 9,9 = 0

X = 0 : 9,9

X = 0

Bài 3 :

Cách 1 :

1,2 x 3,4 : 0,5 = ( 1,2 : 0,5 ) x 3,4 = 2,4 x 3,4 = 8,16

Cách 2 :

1,2 x 3,4 : 0,5 = 4,08 : 0,5 = 8,16

Cách 3 :

1,2 x 3,4 : 0,5 = 1,2 x ( 3,4 : 0,5 ) = 1,2 x 6,8 = 8,16

bài 1:

X x 0,25 + X x 1/0,25=4,59

X x(0,25 + 1/0,25)=4,59

X x [(0,0625+1)/ 0,25]=4,59

X x (1,0625/ 0,25)=4,59

X x 4,25=4,59

X= 4,59 / 4,25

X=1,08

bài 2:

X : 0,1- X x 0,1=-9,9+9,9

X x 1/0,1- X x0,1=0

X x(1/ 0,1 -0,1)=0

X x [(1- 0,01)/ 0,1]=0

X x 0,99/0,1=0

X x 9,9 =0

X=0

Bài 1.

a) Ta có

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

Lại có: 

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

b) \(h\left(x\right)=f\left(x\right)+g\left(x\right)\)

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

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

c) \(h\left(x\right)=0\)

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

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

TH1: \(x=0\)

TH2: \(3x+1=0\) hay \(x=-\dfrac{1}{3}\)

Vậy nghiệm của \(h\left(x\right)\) là \(x=0;x=-\dfrac{1}{3}\)

Bài 2.

a) Ta có \(\left\{{}\begin{matrix}A\left(x\right)=6x^3+5x^2\\B\left(x\right)=x^3-x^2\\C\left(x\right)=-2x^3+4x^2\end{matrix}\right.\)

\(D\left(x\right)=A\left(x\right)+B\left(x\right)-C\left(x\right)\)

\(D\left(x\right)=\left(6x^3+x^3-\left(-2x^3\right)\right)+\left(5x^2-x^2-4x^2\right)\)

\(D\left(x\right)=9x^3\)

b) \(D\left(x\right)=0\)

\(9x^3=0\\ x^3=0\\ x=0\)

Vậy nghiệm của đa thức \(D\left(x\right)\) là \(x=0\).

1

a

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

b

\(=12\left(xy-xz\right)+3x\left(xy-xz\right)\\ =\left(12+3x\right)\left(xy-xz\right)\\ =x\left(12+3x\right)\left(y-z\right)\)

c

\(=\dfrac{1}{2}\left(x^4+2x^2y^2+y^4\right)-\dfrac{1}{2}.4x^2y^2\\ =\dfrac{1}{2}\left(x^4+2x^2y^2+y^4-4x^2y^2\right)\\ =\dfrac{1}{2}\left(x^4-2x^2y^2+y^4\right)\\ =\dfrac{1}{2}\left(x^2-y^2\right)^2\\ =\dfrac{1}{2}\left(x^2-y^2\right)\left(x^2-y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)\left(x+y\right)\left(x-y\right)\left(x+y\right)\)

d

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

2

\(5x^2z-10xyz+5y^2z\\ =5x^2z-5xyz-5xyz+5y^2z\\ =\left(5x^2z-5xyz\right)-\left(5xyz-5y^2z\right)\\ =5xz\left(x-y\right)-5yz\left(x-y\right)\\ =\left(5xz-5yz\right)\left(x-y\right)\\ =5z\left(x-y\right)\left(x-y\right)\\=5z\left(x-y\right)^2\)

Thế \(x=124;y=24;z=2\) vào biểu thức được:

\(5.2\left(124-24\right)^2=10.\left(100\right)^2=10.10000=100000\)