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1) Ta có : \(\frac{2016a+b+c+d}{a}=\frac{a+2016b+c+d}{b}=\frac{a+b+2016c+d}{c}=\frac{a+b+c+2016d}{d}\)
Trừ 4 vế với 2015 ta được : \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Nếu a + b + c + d = 0
=> a + b = -(c + d)
=> b + c = (-a + d)
=> c + d = -(a + b)
=> d + a = (-b + c)
Khi đó M = (-1) + (-1) + (-1) + (-1) = - 4
Nếu a + b + c + d\(\ne0\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)
Khi đó M = 1 + 1 + 1 + 1 = 4
2) a) Ta có : \(\hept{\begin{cases}\left|x+2013\right|\ge0\forall x\\\left(3x-7\right)^{2004}\ge0\forall y\end{cases}\Rightarrow\left|x+2013\right|+\left(3x-7\right)^{2014}\ge0}\)
Dấu "=" xảy ra \(\hept{\begin{cases}x+2013=0\\3y-7=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2013\\y=\frac{7}{3}\end{cases}}}\)
b) 72x + 72x + 3 = 344
=> 72x + 72x.73 = 344
=> 72x.(1 + 73) = 344
=> 72x = 1
=> 72x = 70
=> 2x = 0 => x = 0
c) Ta có :
\(\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{5}{x+4}\Leftrightarrow\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{10}{2x+8}=\frac{7-10}{2x+2-2x-8}=\frac{1}{2}\)(dãy tỉ số bằng nhau)
=> 2x + 2 = 14 => x = 6 ;
2y - 4 = 6 => y = 5 ;
6 + 5 + z = 17 => z = 6
Vậy x = 6 ; y = 5 ; z = 6
3) a) Ta có : \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)(dãy ti số bằng nhau)
=> a + b + c = a + b - c => a + b + c - a - b + c = 0 => 2c = 0 => c = 0;
Lại có : \(\frac{a+b+c}{a+b-c}-1=\frac{a-b+c}{a-b-c}-1\Leftrightarrow\frac{2c}{a+b-c}=\frac{2c}{a-b-c}\Rightarrow a+b-c=a-b-c\) => b = 0
Vậy c = 0 hoặc b = 0
c) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b+b+c+a+c}{c+a+b}=2\)(dãy tỉ số bằng nhau)
=> \(\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)
Khi đó P = \(\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{b}{a}\right)=\frac{b+c}{b}.\frac{c+a}{c}=\frac{a+b}{a}=\frac{2a.2b.2c}{abc}=8\)
Vậy P = 8
2. b) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
\(\frac{-2}{3}\)\(.\)\(x\)\(=\)\(\frac{4}{5}\)
=> \(x\)\(=\)\(\frac{4}{5}\)\(:\)\(\frac{-2}{3}\)
\(x\)\(=\)\(\frac{4}{5}\)\(.\)\(\frac{-3}{2}\)
\(x\)\(=\)\(\frac{-6}{5}\)
Vậy đáp án C đúng
Có: x:y:z=2:3:5
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow x=2k;y=3k;z=5k\)
\(\Rightarrow xyz=2k.5k.3k=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)
=> x=...
y=...
z=...
Có: VT\(\ge0\)( tự xét )
Theo bài ra lại có: VT\(\le0\)
=> VT=0
\(\Rightarrow\hept{\begin{cases}x_1p=y_1q\\.............\\x_mp=y_mq\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x_1}{y_1}=\frac{q}{p}\\...............\\\frac{x_m}{y_m}=\frac{q}{p}\end{cases}}\)
\(\Rightarrow\frac{x_1}{y_1}=\frac{x_2}{y_2}=.....=\frac{x_m}{y_m}=\frac{q}{p}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
........................................................................
những bài khác chốc về làm nốt cho
1) Ta có : Đặt M = 3x + 1 + 3x + 2 + ... + 3x + 100
= 3x(3 + 32 + ... + 3100)
= 3x[(3 + 32 + 33 + 34) + (35 + 36 + 37 + 38) + ... + (397 398 + 399 + 3100)]
= 3x[(3 + 32 + 33 + 34) + 34.(3 + 32 + 33 + 34) + ... + 396.(3 + 32 + 33 + 34)]
= 3x(120 + 34.120 + .... + 396.120)
= 3x.120.(1 + 34 + .... + 396)
=> \(M⋮120\)(ĐPCM)
2) Ta có \(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)
\(\Rightarrow\frac{3a+b+c}{a}-2=\frac{a+3b+c}{b}-2=\frac{a+b+3c}{c}-2\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
Nếu a + b + c = 0
=> a + b = - c
b + c = -a
c + a = -b
Khi đó P = \(\frac{-c}{c}+\frac{-a}{a}+\frac{-b}{b}=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)
Nếu a + b + c \(\ne\)0
=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Khi đó P = \(\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)
Vậy nếu a + b + c = 0 thì P = -3
nếu a + b + c \(\ne\)0 thì P = 6
Ta có :
\(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\)
\(=\left(3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}\right)+...\)\(+\left(3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100}\right)\)
\(=3^x\left(3+3^2+3^3+3^4\right)+...+3^{x+96}\left(3+3^2+3^3+3^4\right)\)
\(=3^x.120+3^{x+4}.120+...+3^{x+96}.120\)
\(=120.\left(3^x+3^{x+4}+...+3^{x+96}\right)\)
Vì \(120⋮120\)
\(\Rightarrow120.\left(3^x+3^{x+4}+...+3^{x+96}\right)⋮120\)
\(\Rightarrow3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}⋮120\left(\forall x\inℕ\right)\left(đpcm\right)\)
\(\frac{5}{4}\cdot\left(-\frac{12}{7}\right)=\frac{-60}{28}=\frac{-15}{7}\)
\(\frac{-4}{3}:\frac{13}{9}=-\frac{4}{3}\cdot\frac{9}{13}=-\frac{36}{39}=-\frac{12}{13}\)
\(-\frac{5}{7}\cdot\frac{49}{3}:\frac{7}{-6}=-\frac{5}{7}\cdot\frac{49}{3}\cdot-\frac{6}{7}=\frac{1470}{147}=10\)
\(-\frac{9}{25}:6=-\frac{9}{150}=-\frac{3}{50}\)
2.
\(\frac{7}{4}:\left(\frac{2}{3}-\frac{5}{4}\right)\cdot\left(-\frac{1}{4}\right)\)
\(=\frac{7}{4}:\left(-\frac{7}{12}\right)\cdot\left(-\frac{1}{4}\right)\)
\(=\frac{3}{4}\)
b,giá trị của x thỏa mãn đẳng thức \(-\frac{2}{3}x=\frac{4}{5}\)là
\(-\frac{2}{3}x=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=-\frac{6}{5}\)
Thay x = 3 vào \(\frac{a-x}{3}=\frac{bx-5}{5}\)
\(\Rightarrow\frac{a-3}{3}=\frac{3b-5}{5}\)\(\Rightarrow\frac{a}{3}-1=\frac{3b}{5}-1\)\(\Rightarrow\frac{a}{3}=\frac{3b}{5}\)\(\Rightarrow a=\frac{3.3b}{5}=\frac{9b}{5}\)
Thay a = 9b/5 vào \(\frac{a}{b}-\frac{b}{a}\)\(\Rightarrow\frac{\frac{9b}{5}}{b}-\frac{b}{\frac{9b}{5}}=\frac{\left(\frac{9b}{5}\right)^2-b^2}{\frac{9b}{5}.b}=\frac{\frac{81b^2}{25}-b^2}{\frac{9b^2}{5}}=b^2\left(\frac{81}{25}-1\right)\div\frac{9b^2}{5}=\frac{56b^2}{25}.\frac{5}{9b^2}=\frac{56}{45}\)
Vậy....