\(\dfrac{1}{2}x=\dfrac{2}{3}y=\dfrac{3}{4}z\)

\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{\dfrac{3}{2}}=\dfrac{z}{\dfrac{4}{3}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2}=\dfrac{y}{\dfrac{3}{2}}=\dfrac{z}{\dfrac{4}{3}}=\dfrac{x-y}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30\)

\(\Rightarrow\left\{{}\begin{matrix}x=30.2=60\\y=30.\dfrac{3}{2}=45\\z=30.\dfrac{4}{3}=40\end{matrix}\right.\)

Coi giá bán xe ô tô ban đầu 100%

Vậy để lãi 30% thì cần bán xe ô đó với giá:

2000000:100x(100+30)=2600000 (đồng)

ô tô gì rẻ thế

\(n^3-n\)

\(=n\left(n^2-1\right)\)

\(=n\left(n+1\right)\left(n-1\right)\)

\(=\left(n-1\right)n\left(n+1\right)\)

Dễ thấy: \(n-1;n;n+1\) là 3 số tự nhiên liên tiếp thì chia hết cho 6

Ta có đpcm

\(A=2x^2+6x\)

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

\(A=2\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)

\(A=2\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{9}{4}.2\)

\(A=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge\dfrac{9}{2}\)

Dấu "=" xảy ra khi :

\(x=-\dfrac{3}{2}\)

B đã sửa đề vì theo đề của you thì ko có tổng nào = nhau

\(B=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(B=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(B=\left[x\left(x+4\right)+1\left(x+4\right)\right]\left[x\left(x+3\right)+2\left(x+3\right)\right]\)

\(B=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)\)

\(B=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

\(B=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)\)

\(B=\left(x^2+5x+5\right)^2-1\ge-1\)

Dấu "=" xảy ra khi:

\(x^2+5x+5=0\)

Cạn....

cau 1 la 46

cau 2 la 10 000 000 dong

cau 3 la 2,8

cau 4 la 39,25kg

cau 5 la 85,2kg

cau 6 la 80

cau 7 la 45 quyen

cau 8 la 200 con 

cau 9 la 15

\(A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)

Vì \(a;b;c\) là các số thực dương nên:

\(\left\{{}\begin{matrix}\dfrac{a}{a+b}>\dfrac{a}{a+b+c}\\\dfrac{b}{b+c}>\dfrac{b}{a+b+c}\\\dfrac{c}{c+a}>\dfrac{c}{a+b+c}\end{matrix}\right.\)

Cộng theo 3 vế :

\(A>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\)(1)

Vì \(a;b;c\) là 3 số thực dương nên \(\dfrac{a}{a+b};\dfrac{b}{b+c};\dfrac{c}{c+a}< 1\) nên:

\(\left\{{}\begin{matrix}\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\\\dfrac{b}{b+c}< \dfrac{a+b}{a+b+c}\\\dfrac{c}{c+a}< \dfrac{b+c}{a+b+c}\end{matrix}\right.\)

Cộng theo 3 vế:

\(A< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}=2\)(2)

Từ (1) và (2) ta có:

\(1< A< 2\)

Ta có:\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow100^o+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow\widehat{B}+\widehat{C}=80^o\)

Nên \(\left\{{}\begin{matrix}\widehat{B}-\widehat{C}=50^o\\\widehat{B}+\widehat{C}=80^o\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}+\widehat{C}+\widehat{B}-\widehat{C}=50^o+80^o\\\widehat{B}+\widehat{C}-\widehat{B}+\widehat{C}=80^o-50^o\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2\widehat{B}=130^o\\2\widehat{C}=30^o\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=65^o\\\widehat{C}=15^o\end{matrix}\right.\)

\(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-....-\dfrac{1}{1024}\)

\(A=-1-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
Đặt:

\(B=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(2B=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(B=1-\dfrac{1}{2^{10}}\)

Thay vào \(A\)

\(A=-1-\left(1-\dfrac{1}{2^{10}}\right)\)

\(A=-1-1+\dfrac{1}{2^{10}}\)

\(A=-2+\dfrac{1}{2^{10}}\)

Mk nghĩ là:

\(\left\{{}\begin{matrix}\sqrt{17}>\sqrt{16}=4\\\sqrt{26}>\sqrt{25}=5\end{matrix}\right.\) \(\Rightarrow\sqrt{17}+\sqrt{26}+1>4+5+1=10\)

\(\sqrt{99}< \sqrt{100}=10\)

Nên: \(\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)

\(A=\dfrac{1}{1-\dfrac{1}{1-2^{-1}}}+\dfrac{1}{1+\dfrac{1}{1+2^{-1}}}\)

\(A=\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{2}}}+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{2}}}\)

\(A=\dfrac{1}{1-\dfrac{1}{\dfrac{1}{2}}}+\dfrac{1}{1+\dfrac{1}{\dfrac{3}{2}}}\)

\(A=\dfrac{1}{1-2}+\dfrac{1}{1+\dfrac{2}{3}}\)

\(A=\dfrac{1}{-1}+\dfrac{1}{\dfrac{5}{3}}\)

\(A=-1+\dfrac{3}{5}\)

\(A=-\dfrac{2}{5}\)