Đại số lớp 7 - Hỏi đáp

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)(T/C...)

Xét a+b+c=0

\(\Rightarrow a+b=-c,c+b=-a,a+c=-b\)

\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)

Xét a+b+c\(\ne0\)

\(\Rightarrow a+b=2c,b+c=2a,c+a=2b\)

\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=\frac{2c}{b}\cdot\frac{2a}{c}\cdot\frac{2b}{a}=8\)

Giải:
+) Xét a + b + c = 0

\(\Rightarrow-a=b+c\)

\(\Rightarrow-b=a+c\)

\(\Rightarrow-c=a+b\)

Ta có:

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{-c}{c}=\frac{-a}{a}=\frac{-b}{b}=-1\)

Lại có: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=-1\)

+) Xét \(a+b+c\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

Ta có:

\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=2.2.2=8\)

Vậy M = -1 hoặc M = 8

Từ \(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)

\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)

\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{c}{ca}+\dfrac{a}{ca}\)

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{b}+\dfrac{1}{a}=\dfrac{1}{c}+\dfrac{1}{b}\\\dfrac{1}{c}+\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\\\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{1}{b}+\dfrac{1}{a}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\\\dfrac{1}{c}=\dfrac{1}{b}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\Rightarrow a=b=c\)

Khi đó: \(M=\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{1\cdot1+1\cdot1+1\cdot1}{1^2+1^2+1^2}=\dfrac{3}{3}=1\)

Gọi x là số cần tìm và a,b,c, lần lượt là các số của nó (x thuộc N*)

Nếu x chia hết cho 18 suy ra x chia hết cho 2 nên x chẵn

Ta có : a,b,c, tỉ lệ với 1:2:3 thì nhân theo hệ quả ta được các số 123 ; 246 ; 369

Mà x chia hết cho 9 suy ra x chia hết cho 3

Thỏa mãn các điều kiện trên ta được các số 396 và 936

Vì x chia hết cho 18 suy ra x = 936

Vậy số cần tìm là 936.

Ta có: 2xy + x - 2y = 4

=> 2y(x - 1) + x = 4

=> 2y(x - 1) + x - 1 = 3

=> 2y(x - 1) + (x - 1) = 3

=> (x - 1).(2y + 1) = 3

=> x-1 và 2y+1 là Ư(3)={-3;-1;1;3}

Ta có bảng:

Vậy ...

2xy+x-2y = (2y+1)x - 2y =(2y+1)x -(2y+1) +1=(x-1)(2y+1)+1

vì 2xy+x-2y = 4

=>

(x-1)(2y+1)= 4 -1 =3

vậy x,y bằng trên kia có kìa hihi.

a) \(4^x+4^{x+3}=4160\)

\(\Rightarrow4^x+4^x.4^3=4160\)

\(\Rightarrow4^x.\left(1+4^3\right)=4160\)

\(\Rightarrow4^x.65=4160\)

\(\Rightarrow4^x=64\)

\(\Rightarrow4^x=4^4\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

b) \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(\Rightarrow2^x.\frac{1}{2}+5.2^x.\frac{1}{4}=\frac{7}{32}\)

\(\Rightarrow2^x.\left(\frac{1}{2}+5.\frac{1}{4}\right)=\frac{7}{32}\)

\(\Rightarrow2^x.\frac{7}{4}=\frac{7}{32}\)

\(\Rightarrow2^x=\frac{7}{32}:\frac{7}{4}\)

\(\Rightarrow2^x=\frac{1}{8}\)

\(\Rightarrow2^x=2^{-3}\)

\(\Rightarrow x=-3\)

Vậy \(x=-3\)

b)\(2^{x-1}+5\cdot2^{x-2}=\frac{7}{32}\)

\(2^x:2+5\cdot2^x:2^2=\frac{7}{32}\)

\(2^x:2+2^x:\frac{4}{5}=\frac{7}{32}\)

\(2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\)

\(2^x\cdot\frac{7}{4}=\frac{7}{32}\)

\(2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}\)

\(2^x=\frac{2^0}{2^3}=2^{-3}\)

\(\Rightarrow x=-3\)

\(2n-3⋮n+1\)

\(\Rightarrow2n+2-5⋮n+1\)

\(\Rightarrow2\left(n+1\right)-5⋮n+1\)

\(\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\inƯ\left(5\right)\)

\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}n+1=1\Rightarrow n=0\\n+1=-1\Rightarrow n=-2\\n+1=5\Rightarrow n=4\\n+1=-5\Rightarrow n=-6\end{matrix}\right.\)

2n-3 chia hết cho n+1

=>2n+2-5 chia hết cho n+1

=>2(n+1)-5 chia hết cho n+1

Mà 2(n+1) chia hết cho n+1=>5 chia hết cho n+1

=>n+1 thuộc Ư(5)={1;-1;5;-5}

TH1:n+1=1,=>n=0 thuộc Z

TH2:n+1=-1=>n=-2 thuộc Z

TH3:n+1=5=>n=4 thuộc Z

TH4:n+1=-5=>n=-6 thuộc Z

Vậy n thuộc {0;-2;4;-6}

Chúc Bạn Học Tốt !!!

Lời giải:

Có \(f(0),f(1),f(2)\in\mathbb{Z}\Rightarrow \left\{\begin{matrix} f(0)=c\in\mathbb{Z}\\ f(1)=a+b+c\in\mathbb{Z}\\ f(2)=4a+2b+c\in\mathbb{Z}\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} a+b\in\mathbb{Z}\\ 4a+2b\in\mathbb{Z}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2a+2b\in\mathbb{Z}\\ 4a+2b\in\mathbb{Z}\end{matrix}\right.\Rightarrow 2a\in\mathbb{Z}\rightarrow 2b\in\mathbb{Z}\)

Ta có đpcm

Ta có: \(2x+\dfrac{1}{7}=\dfrac{1}{y}\)

\(\Rightarrow\dfrac{14x}{7}+\dfrac{1}{7}=\dfrac{1}{y}\)

\(\Rightarrow\dfrac{14x+1}{7}=\dfrac{1}{y}\)

\(\Rightarrow\left(14x+1\right)y=7\)

Vì \(x,y\in Z\Rightarrow\left[{}\begin{matrix}14x+1\in Z\\y\in Z\end{matrix}\right.\)

\(\Rightarrow14x+1\inƯ\left(7\right);y\inƯ\left(7\right)\)

Từ đó lập bảng để xét các t/h nhé!

Vậy \(\left(x,y\right)\in\left\{\left(...,...\right);...\right\}\)

x ϵ z

nhưng x là số gì

\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)

\(\Rightarrow2A=2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\)

\(\Rightarrow2A-A=\left(2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)\)

\(\Rightarrow A=\left(2-1\right)+\frac{3}{2^2}+\left(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)

Mà \(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}=\frac{1}{2^2}-\frac{1}{2^{99}}\) ( tự tính )

\(\Rightarrow A=1+\frac{3}{2^2}+\frac{1}{2^2}-\frac{1}{2^{99}}-\frac{100}{2^{100}}\)

\(\Rightarrow A=1+1-\frac{1}{2^{99}}-\frac{100}{2^{100}}\)

\(\Rightarrow A=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}\)

Vậy...

=\(\frac{1}{2}+\frac{2}{2^3}+\frac{3}{2^3}+....+\frac{100}{2^{100}}\)

=\(\frac{1}{2}+\frac{2}{2^3}+\frac{3}{2^3}+....+\frac{100}{2^{100}}\)+\(\frac{1}{2}\left(\frac{1}{2}+\frac{2}{2^2}+...+\frac{99}{2^{99}}\right)\)

= 1-\(\frac{1}{2^{100}}+\frac{1}{2}\left(A-\frac{100}{2^{100}}\right)\)

<=> \(\frac{1}{2}A=1-\frac{1}{2^{100}}-\frac{100}{2^{100}}=1-\frac{101}{2^{100}}\)

=\(\frac{2^{100}-101}{2^{100}}\Rightarrow A=2\)

Các bn cố giải giùm mk vs

Có: \(\frac{a}{b}\left(a,b\in N,b\ne0\right)\)

\(\frac{a}{b}>\frac{4}{7}\)

\(\Rightarrow7a>4b\)

\(\Leftrightarrow8b< 1994\)

\(\Leftrightarrow b< 249\)

\(7a>4b\)

\(\Leftrightarrow14a>1994\)

\(\Leftrightarrow a>142\)

Có: \(\frac{a}{b}< \frac{2}{3}\)

\(\Rightarrow3a< 2b\)

\(\Leftrightarrow6a+7a< 4b+7a\)

\(\Leftrightarrow13a< 1994\)

\(\Leftrightarrow a< 154\)

Có:\(3a< 2b\)

\(\Leftrightarrow6a+a+4b< 8b+a\)

\(\Leftrightarrow1994< 8b+a\)

mà a=\(\frac{1994-4b}{7}\)

\(8b+a=8b+\frac{1994-4b}{7}>1994\)

\(\Leftrightarrow56b+1994-4b>13958\)

\(\Leftrightarrow b>230\)

Vậy \(\frac{4}{7}< \frac{a}{b}< \frac{2}{3}\Leftrightarrow a,b\in N;142< a< 154;230< b< 249\)

Nguyễn Việt Lâm Bài này có cần tìm cụ thể ko?

Có\(\left\{{}\begin{matrix}\left|x-1\right|\ge x-1\\\left|x-2\right|\ge x-2\\\left|x-3\right|\le x-3\\\left|x-4\right|\le x-4\end{matrix}\right.\forall x\)

\(\Rightarrow A=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\ge\left(x-1\right)+\left(x-2\right)+\left(3-x\right)+\left(4-x\right)\)\(\Rightarrow A\ge4\)

Dấu "=" xảy ra khi\(\left\{{}\begin{matrix}x-1\ge0\\x-2\ge0\\x-3\le0\\x-4\le0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\ge2\\x\le3\\x\le4\end{matrix}\right.\Rightarrow2\le x\le4\)

Mà x\(\in Z\)\(\Rightarrow x=2;x=3\)

Vậy với x\(\in\left\{2;3\right\}\)thì A đạt giá trị nhỏ nhất là 4

Ta có: \(\dfrac{1+2a}{15}=\dfrac{7-3a}{20}\Rightarrow20\left(1+2a\right)=15\left(7-3a\right)\Leftrightarrow20+40a=105-45a\Leftrightarrow40a+45a=105-20\Leftrightarrow85a=85\Leftrightarrow a=1\)

Ta có: \(\dfrac{7-3a}{20}=\dfrac{3b}{23+7a}\Rightarrow\dfrac{7-3.1}{20}=\dfrac{3b}{23+7.1}\Rightarrow\dfrac{4}{20}=\dfrac{3b}{30}\Rightarrow\dfrac{1}{5}=\dfrac{b}{10}\Rightarrow b=2\) Vậy a=1;b=2

Ta có:

\(\left|x+1\right|\ge0,\left|x+2\right|\ge0,...,\left|x+2014\right|\ge0\)

\(\Rightarrow\)\(\left|x+1\right|+\left|x+2\right|+...+\left|x+2014\right|\ge0\)

\(\Rightarrow2015x\ge0\)

\(\Rightarrow x\ge0\)

Khi đó :\(\left|x+1\right|=x+1,\left|x+2\right|=x+2,...,\left|x+2014\right|=x+2014\)\(\Rightarrow x+1+x+2+...+x+2014=2015x\)

\(\Rightarrow2014x+1+2+...+2014=2015x\)

\(\Rightarrow1+2+..+2014=x\)

\(\Rightarrow x=\dfrac{\left(1+2014\right)2014}{2}=2029105\)