mai mk giúp cho. hôm nay mik bận làm đề cương rồi

Bài 1:

Vận tốc cano khi dòng nước lặng là: $25-2=23$ (km/h) 

Bài 2:

Đổi 1 giờ 48 phút = 1,8 giờ

Độ dài quãng đường AB: $1,8\times 25=45$ (km) 

Vận tốc ngược dòng là: $25-2,5-2,5=20$ (km/h) 

Cano ngược dòng từ B về A hết:

$45:20=2,25$ giờ = 2 giờ 15 phút.

(ĐÂY CHỈ LÀ CÁCH CỦA MÌNH THÔI NHA)

d)

Gọi x là độ dài của MN.

Ta có: AH = AK + KH (gt)

=> KH = AH -AK

hay KH = 9,6-3,6 =6

Ta có: S_ABC = S_AMN + S_MNBC (gt)

hay \(\dfrac{AK.MN}{2}+\dfrac{KH\left(BC+MN\right)}{2}\) = \(\dfrac{AB.AC}{2}\)

hay \(\dfrac{3,6.x}{2}+\dfrac{6\left(x+20\right)}{2}=\dfrac{12.16}{2}=96\)

\(\Leftrightarrow\) 3,6x + 6x + 120 = 96.2 = 192

\(\Leftrightarrow\) 9.6x = 192 - 120= 72

\(\Leftrightarrow\) x = \(\dfrac{72}{9,6}=7,5\)

S_MNCB= \(\dfrac{KH\left(MN.BC\right)}{2}=\dfrac{6\left(7,5+20\right)}{2}=82,5\) (cm²)

B A C H 20 12 16 k AK=6 AH=9,6 M N MN // BC

đề giống bọn mk này

bạn tính diện tích ABC xong trừ đi diện tích AMN là ra kết quả là 82,5

Điều kiện:

\(x-1\ne0\Rightarrow x\ne1\)

\(x^3+x\ne0\Leftrightarrow x\ne0\)

Bài 1:

a.

$a^3-a^2c+a^2b-abc=a^2(a-c)+ab(a-c)$

$=(a-c)(a^2+ab)=(a-c)a(a+b)=a(a-c)(a+b)$

b.

$(x^2+1)^2-4x^2=(x^2+1)^2-(2x)^2=(x^2+1-2x)(x^2+1+2x)$

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

c.

$x^2-10x-9y^2+25=(x^2-10x+25)-9y^2$

$=(x-5)^2-(3y)^2=(x-5-3y)(x-5+3y)$

d.

$4x^2-36x+56=4(x^2-9x+14)=4(x^2-2x-7x+14)$

$=4[x(x-2)-7(x-2)]=4(x-2)(x-7)$

Bài 2:

a. $(3x+4)^2-(3x-1)(3x+1)=49$

$\Leftrightarrow (3x+4)^2-[(3x)^2-1]=49$

$\Leftrightarrow (3x+4)^2-(3x)^2=48$

$\Leftrightarrow (3x+4-3x)(3x+4+3x)=48$

$\Leftrightarrow 4(6x+4)=48$

$\Leftrightarrow 6x+4=12$

$\Leftrightarrow 6x=8$

$\Leftrightarrow x=\frac{4}{3}$

b. $x^2-4x+4=9(x-2)$

$\Leftrightarrow (x-2)^2=9(x-2)$

$\Leftrightarrow (x-2)(x-2-9)=0$

$\Leftrightarrow (x-2)(x-11)=0$

$\Leftrightarrow x-2=0$ hoặc $x-11=0$

$\Leftrightarrow x=2$ hoặc $x=11$

c.

$x^2-25=3x-15$

$\Leftrightarrow (x-5)(x+5)=3(x-5)$

$\Leftrightarrow (x-5)(x+5-3)=0$

$\Leftrightarrow (x-5)(x+2)=0$

$\Leftrightarrow x-5=0$ hoặc $x+2=0$

$\Leftrightarrow x=5$ hoặc $x=-2$

\(\left(n^2-1\right)^{2016}:n\)

Ta có \(n^2⋮n\)

\(\Rightarrow\left(n^2\right)^{2016}⋮n\)

Mà \(\left(-1\right)^{2016}:n=a\left(dư1\right)\)

Vậy số dư khi chia \(\left(n^2-1\right)^{2016}\) cho \(n\) là 1.

a)tam giác BHA có BI là phân giác(góc ABI=góc HBI) nên \(\dfrac{AI}{IH}=\dfrac{AB}{BH}\Rightarrow AI\cdot BH=AB\cdot IH\)

b)xét tam giác BHA và tam giác BAC có:

góc ABC chung

góc BHA=góc BAC=90 độ

\(\Rightarrow\Delta BHA\infty\Delta BAC\left(g.g\right)\\ \Rightarrow\dfrac{BH}{AB}=\dfrac{AB}{BC}\Rightarrow AB^2=BH\cdot BC\)

c)ta có:

theo câu a) \(\dfrac{AI}{IH}=\dfrac{AB}{BH}\Rightarrow\dfrac{IH}{AI}=\dfrac{BH}{AB}\left(1\right)\)

theo câu b) \(\dfrac{BH}{AB}=\dfrac{AB}{BC}\)

ta lại có BD là phân giác góc ABC nên \(\dfrac{AB}{BC}=\dfrac{AD}{DC}\Rightarrow\dfrac{AD}{DC}=\dfrac{BH}{AB}\)(2)

từ (1) và (2)\(\Rightarrow\dfrac{IH}{IA}=\dfrac{AD}{DC}\left(=\dfrac{BH}{AB}\right)\)

cảm ơn bn

33.

\(x^{10}+x^5+1\\ =x^{10}+x^9+x^8-x^9-x^8-x^7+x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\\ =x^8\left(x^2+x+1\right)-x^7\left(x^2+x+1\right)+x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ \left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)

34.

đặt: \(t=x^2+x+1,5\)

khi đó:

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\\ =\left(t-0,5\right)\left(t+0,5\right)-12\\ =t^2-0,25-12\\ =t^2-12,25\\ =\left(t-3,5\right)\left(t+3,5\right)\\ =\left(x^2+x-2\right)\left(x^2+x+5\right)\)

35.

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+1\\ =\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\\ =\left(x^2-5x+5-1\right)\left(x^2-5x+5+1\right)+1\\ =\left(x^2-5x+5\right)^2-1+1\\ =\left(x^2-5x+5\right)^2\)

36.

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+15\\ =\left(x^2-10x+16\right)\left(x^2-10x+24\right)+15\\ =\left(x^2-10x+20-4\right)\left(x^2-10x+20+4\right)+15\\ =\left(x^2-10x+20\right)^2-4^2+15\\ =\left(x^2-10x+20\right)^2-1\\ =\left(x^2-10x+19\right)\left(x^2-10x+21\right)\)

37.

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\\ =\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\\ =\left(x^2-10x+20-4\right)\left(x^2-10x+20+4\right)+16\\ =\left(x^2-10x+20\right)^2-4^2+16\\ =\left(x^2-10x+20\right)^2\)

38.

\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)-24\\ =\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\\ =\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\\ =\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)-24\\ =\left(x^2+5x+5\right)^2-1-24\\ =\left(x^2+5x+5\right)^2-5^2\\ =\left(x^2+5x+10\right)\left(x^2+5x\right)\\ =x\left(x+5\right)\left(x^2+5x+10\right)\)

39.

\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\\ =\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\\ =\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\\ =\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)+1\\ =\left(x^2+5x+5\right)^2-1+1\\ =\left(x^2+5x+5\right)^2\)

40.

\(a^2b^2\left(a-b\right)-c^2b^2\left(c-b\right)+a^2c^2\left(c-a\right)\\ =a^3b^2-a^2b^3-c^3b^2+c^2b^3+a^2c^2\left(c-a\right)\\ =b^2\left(a^3-c^3\right)+b^3\left(c^2-a^2\right)+a^2c^2\left(c-a\right)\\ =b^2\left(a-c\right)\left(a^2+ac+c^2\right)+b^3\left(c-a\right)\left(c+a\right)+a^2c^2\left(c-a\right)\\ =-b^2\left(c-a\right)\left(a^2+ac+c^2\right)+\left(c-a\right)\left(cb^3+ab^3+a^2c^2\right)\\ =\left(c-a\right)\left(cb^3+ab^3+a^2c^2-a^2b^2-acb^2-b^2c^2\right)\)

42.

\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\\ =ab^2-a^2b-b^2c+bc^2-ac\left(c-a\right)\\ =b^2\left(a-c\right)+b\left(c^2-a^2\right)-ac\left(c-a\right)\\ =\left(a-c\right)\left(b^2-ac+ba+bc\right)\)

Giup cai j ? Cau nao ?

Đề số 3.

1.

a,\(4x\left(5x^2-2x+3\right)\)

\(=20x^3-8x^2+12x\)

b.\(\left(x-2\right)\left(x^2-3x+5\right)\)

\(=x^3-3x^2+5x-2x^2+6x-10\)

\(=x^3-5x^2+11x-10\)

c,\(\left(10x^4-5x^3+3x^2\right):5x^2\)

\(=2x^2-x+\dfrac{3}{5}\)

d,\(\left(x^2-12xy+36y^2\right):\left(x-6y\right)\)

\(=\left(x-6y\right)^2:\left(x-6y\right)\)

\(=x-6y\)

2.

a,\(x^2+5x+5xy+25y\)

\(=\left(x^2+5x\right)+\left(5xy+25y\right)\)

\(=x\left(x+5\right)+5y\left(x+5\right)\)

\(=\left(x+5y\right)\left(x+5\right)\)

b,\(x^2-y^2+14x+49\)

\(=\left(x^2+14x+49\right)-y^2\)

\(=\left(x+7\right)^2-y^2\)

\(=\left(x+7-y\right)\left(x+7+y\right)\)

c,\(x^2-24x-25\)

\(=x^2+25x-x-25\)

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

\(=x\left(x-1\right)+25\left(x-1\right)\)

\(=\left(x+25\right)\left(x-1\right)\)

3.

a,\(5x\left(x-3\right)-x+3=0\)

\(5x\left(x-3\right)-\left(x-3\right)=0\)

\(\left(5x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{5}\) hoặc \(x=3\)

b.\(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)

\(3x^2-15x-\left(2x+3x^2-2-3x\right)=30\)

\(3x^2-15x-2x-3x^2+2+3x=30\)

\(-14x+2=30\)

\(-14x=28\)

\(x=-2\)

c,\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)

\(x^2+3x+2x+6-\left(x^2+5x-2x-10\right)=0\)

\(x^2+5x+6-x^2-5x+2x+10=0\)

\(2x+16=0\)

\(2x=-16\)

\(x=-8\)

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