M=4 nha
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tks bn

Bài 1:

\(3^{-1}.3^n+4.3^n=13.3^5\)

\(\Rightarrow3^{n-1}+4.3.3^{n-1}=13.3^5\)

\(\Rightarrow3^{n-1}\left(1+4.3\right)=13.3^5\)

\(\Rightarrow3^{n-1}.13=13.3^5\)

\(\Rightarrow3^{n-1}=3^5\)

\(\Rightarrow n-1=5\)

\(\Rightarrow n=6\)

Vậy n = 6

Bài 2a: Câu hỏi của Nguyễn Trọng Phúc - Toán lớp 7 | Học trực tuyến

4.a

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

Thanks

TH1:a+b+c=0

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(\Rightarrow H=\dfrac{b+a}{b}.\dfrac{c+b}{c}.\dfrac{a+c}{a}=\dfrac{\left(-c\right)\left(-b\right)\left(-a\right)}{b.c.a}=-1\)

TH2:\(a+b+c\ne0\)

Áp dụng tc dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

\(\Rightarrow H=\dfrac{b+a}{b}.\dfrac{c+b}{c}.\dfrac{a+c}{a}=\dfrac{\left(2c\right)\left(2b\right)\left(2a\right)}{b.c.a}=8\)

Vậy H=-1 hoặc H=8

c)

Ta có \(a< b< c< d< m< n\)

\(\Rightarrow\left\{{}\begin{matrix}a< b\\c< d\\m< n\end{matrix}\right.\)

\(\Rightarrow a+c+m\le b+d+n\)

\(\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{1}{2}\)

\(\Leftrightarrow2a+2c+2m< a+b+c+d+m+n\)

\(\Leftrightarrow a+c+m< b+d+n\) ( thỏa mãn đề bài )

\(\Rightarrow\) đpcm

a: \(\dfrac{2}{3}:\left(6x+7\right)=0.2:1\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{2}{3}:\left(6x+7\right)=\dfrac{1}{5}:\dfrac{7}{6}=\dfrac{6}{35}\)

\(\Leftrightarrow6x+7=\dfrac{35}{9}\)

=>6x=-28/9

hay x=-28/54=-14/27

b: \(\dfrac{a}{a+2b}=\dfrac{c}{c+2d}\)

\(\Leftrightarrow a\left(c+2d\right)=c\left(a+2b\right)\)

\(\Leftrightarrow ac+2ad=ac+2bc\)

=>2ad=2bc

=>ad=bc

=>a/b=c/d

Đặt a/b=c/d=k

=>a=bk; c=dk

\(A=\dfrac{a^2\cdot d^2-4b^2\cdot c^2}{abcd}=\dfrac{b^2k^2\cdot d^2-4\cdot b^2\cdot d^2k^2}{bk\cdot b\cdot dk\cdot d}\)

\(=\dfrac{-3b^2k^2d^2}{b^2k^2d^2}=-3\)

Câu 1:

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow\left(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}\right)\) - \(\left(\dfrac{x+1}{13}+\dfrac{x+1}{14}\right)=0\)

\(\Rightarrow\left(x+1\right).\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)\)= 0

Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)

\(\Rightarrow x+1=0\)

=> x = 0 - 1

=> x = -1

Câu 2:

Ta có: \(A=\dfrac{3n+9}{n-4}=\dfrac{3n-3.4+9+12}{n-4}\)

\(=\dfrac{3.\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)

Để A có giá trị nguyên thì:

n - 4 \(\in\) Ư(21)

=> n - 4 \(\in\)

1. Câu hỏi của Cuber Việt ( Câu b í -.- )

2. Quy đồng mẫu số:

\(\dfrac{a}{b}=\dfrac{a.\left(b+2018\right)}{b.\left(b+2018\right)}=\dfrac{ab+2018a}{b.\left(b+2018\right)}\)

\(\dfrac{a+2018}{b+2018}=\dfrac{\left(a+2018\right).b}{\left(b+2018\right).b}=\dfrac{ab+2018b}{b.\left(b+2018\right)}\)

Vì \(b>0\) \(\Rightarrow\) Mẫu 2 phân số ở trên dương.

So sánh \(ab+2018a\) và \(ab+2018b\):

. Nếu \(a< b\Rightarrow\) Tử số phân số thứ 1 < Tử số phân số thứ 2.

\(\Rightarrow\dfrac{a}{b}< \dfrac{a+2018}{b+2018}\)

. Nếu \(a=b\) \(\Rightarrow\) Hai phân số bằng 1.

. Nếu \(a>b\Rightarrow\) Tử số phân số thứ 1 > Tử số phân số thứ 2.

\(\Rightarrow\dfrac{a}{b}< \dfrac{a+2018}{b+2018}\)

3. \(\dfrac{x}{6}-\dfrac{1}{y}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{y}=\dfrac{x}{6}-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{y}=\dfrac{x-3}{6}\)

\(\Rightarrow y.\left(x-3\right)=6\)

Ta có: \(6=1.6=2.3=(-1).(-6)=(-2).(-3)\)

Tự lập bảng ...

Vậy ta có những cặp x,y thỏa mãn là:

\(\left(1,7\right);\left(6,2\right);\left(2,4\right);\left(3,3\right);\left(-1,-5\right);\left(-6,0\right);\left(-2,-2\right);\left(-3,-1\right)\)

\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{a\left(b+2018\right)}{b\left(b+2018\right)}\\\dfrac{a+2018}{b+2018}=\dfrac{b\left(a+2018\right)}{b\left(b+2018\right)}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{ab+2018a}{b^2+2018b}\\\dfrac{a+2018}{b+2018}=\dfrac{ab+2018b}{b^2+2018b}\end{matrix}\right.\)

Cần so sánh:

\(ab+2018a\) với \(ab+2018b\)

Cần so sánh \(2018a\) với \(2018b\)

Cần so sánh \(a\) với \(b\)

\(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2018}{b+2018}\)

\(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2018}{b+2018}\)

\(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2018}{b+2018}\)