Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) 

a) \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{dk}{dk+d}=\dfrac{c}{c+d}\) 

b) \(\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{2c+5d}{3c-4d}\)

c) \(\dfrac{2018a-2019b}{2019c+2020d}=\dfrac{2018bk-2019b}{2019dk+2020d}=\dfrac{b\left(2018k-2019\right)}{d\left(2019k+2020\right)}=\dfrac{b}{d}\cdot\dfrac{2018k-2019}{2019k+2020}\) (1) 

Mà: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{b}{d}=\dfrac{c}{a}\) 

\(\left(1\right)=\dfrac{c}{a}\cdot\dfrac{2018k-2019}{2019k+2020}=\dfrac{2018ck-2019c}{2019ak+2020a}=\dfrac{2018ck-2019dk}{2019ak+2020bk}\\ =\dfrac{k\left(2018c-2019d\right)}{k\left(2019a+2020b\right)}=\dfrac{2018c-2019d}{2019a+2020b}\)

(5.3^5) + 17.3^4) : 6^2

= (5.243 + 17.81) : 36

= (1215 + 1377) : 36

= 2592 : 36

= 72

\(x^2-64\cdot0,5\cdot x+3=0\\ \Leftrightarrow x^2-32x+256=253\\ \Leftrightarrow x^2-32x+16^2=253\\ \Leftrightarrow\left(x-16\right)^2=253\\ \Leftrightarrow\left[{}\begin{matrix}x-16=\sqrt{253}\\x-16=-\sqrt{253}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{253}+16\\x=-\sqrt{253}+16\end{matrix}\right.\)

Vậy...

biện pháp tư từ: "Có những bông hoá lớn và cũng có những bông hoa nhỏ....ở bên vệ đường

sos nha

ai nhanh tick cho

a: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

=>OA+AB=OB

=>AB+2=6

=>AB=4(cm)

b: M là trung điểm của OB

=>\(OM=BM=\dfrac{OB}{2}=3\left(cm\right)\)

Trên tia Ox, ta có: OA<OM(2cm<3cm)

nên A nằm giữaO và M

=>OA+AM=OM

=>AM+2=3

=>AM=1(cm)

a) OA < OB => A nằm giữa O và B 

=> AB = OB - OA = 6 - 2 = 4(cm) 

b) M là trung điểm của OB =>` OM=MB=1/2OB=1/2*6=3(cm) `

=> AM=AB - MB = 4 - 3 = 1(cm)

M(x)=P(x)-Q(x)

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

\(=-8x^5+2x^3+5x^2-x+3\)

\(M\left(-1\right)=-8\left(-1\right)^5+2\cdot\left(-1\right)^3+5\cdot\left(-1\right)^2-\left(-1\right)+3\)

=8-2+5+1+3

=6+6+3

=15

a: Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{B}+\widehat{C}=180^0-75^0=105^0\)

mà \(\widehat{B}-\widehat{C}=15^0\)

nên \(\widehat{B}=\dfrac{105^0+15^0}{2}=60^0;\widehat{C}=60^0-15^0=45^0\)

b: BD là phân giác của góc ABC

=>\(\widehat{ABD}=\dfrac{60^0}{2}=30^0\)

Xét ΔABD có \(\widehat{ABD}+\widehat{BAD}+\widehat{ADB}=180^0\)

=>\(\widehat{ADB}+30^0+75^0=180^0\)

=>\(\widehat{ADB}=75^0\)

t yêu m

Bài 9:

c:Đặt \(\left(x-4\right)\left(x-1\right)\left(x+2\right)^2=0\)

=>\(\left[\begin{array}{l}x-4=0\\ x-1=0\\ \left(x+2\right)^2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=1\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=1\\ x=-2\end{array}\right.\)

k: Đặt \(x^2-4x+3=0\)

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

=>x(x-1)-3(x-1)=0

=>(x-1)(x-3)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=3\end{array}\right.\)

l:Sửa đề: \(2x^2+6x-x^2\left(x+3\right)\)

Đặt \(2x^2+6x-x^2\left(x+3\right)=0\)

=>\(2x\left(x+3\right)-x^2\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(2x-x^2\right)=0\)

=>x(2-x)(x+3)=0

=>\(\left[\begin{array}{l}x=0\\ 2-x=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\\ x=-3\end{array}\right.\)