Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a\left(a+b+c\right)=-12\)
\(b\left(a+b+c\right)=18\)
\(c\left(a+b+c\right)=30\)
\(a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=-12+18+30\)
\(\left(a+b+c\right)\left(a+b+c\right)=36\)
\(\left(a+b+c\right)^2=\left(\pm6\right)^2\)
\(a+b+c=\pm6\)
Th1:
\(a+b+c=6\)
\(\left[\begin{array}{nghiempt}a\times6=-12\\b\times6=18\\c\times6=30\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=-\frac{12}{6}\\b=\frac{18}{6}\\c=\frac{30}{6}\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=-2\\b=3\\c=5\end{array}\right.\)
Th2:
\(a+b+c=-6\)
\(\left[\begin{array}{nghiempt}a\times\left(-6\right)=-12\\b\times\left(-6\right)=18\\c\times\left(-6\right)=30\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=\frac{-12}{-6}\\b=\frac{18}{-6}\\c=\frac{30}{-6}\end{array}\right.\)
\(\left[\begin{array}{nghiempt}a=2\\b=-3\\c=-5\end{array}\right.\)
\(\frac{a}{2}=\frac{b}{3}\rightarrow\frac{a}{6}=\frac{b}{12};\frac{b}{4}=\frac{c}{5}\rightarrow\frac{b}{12}=\frac{c}{20}\)
Ta có: \(\frac{a}{6}=\frac{b}{12}=\frac{c}{20}\) và a+b-c=3
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU:
\(\frac{a}{6}=\frac{b}{12}=\frac{c}{20}=\frac{a+b-c}{6+12-20}=\frac{3}{-2}\)
*\(\frac{a}{6}=-\frac{3}{2}\rightarrow a=6\cdot-\frac{3}{2}=-9\)
*\(\frac{b}{12}=-\frac{3}{2}\rightarrow b=12\cdot-\frac{3}{2}=-18\)
*\(\frac{c}{20}=-\frac{3}{2}\rightarrow c=20\cdot-\frac{3}{2}=-30\)
\(\Leftrightarrow a+b+c=-8+-18+-30=-56\)
dễ mik dùng cácg rút nha!! ta rút b
=> Có biểu thức (2 x b : 3) + b + (5 x b : 4) = 21
nhấn máy tính CASIO => b =7,2
a = (2 x b :3) = (2 x 7,2 :3) =4,8
c= (5 x b :4) = (5 x 7,2 :4)=9
=> 3a-b+c=3 x 4,8 - 7,2 + 9=16,2
Bn k cho mik nha!!!
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) \(7^{2x}+7^{2x+3}=344\)
\(7^{2x}\cdot\left(1+7^3\right)=344\)
\(7^{2x}\cdot\left(1+343\right)=344\)
\(7^{2x}\cdot344=344\)
\(7^{2x}=1\)
\(7^{2x}=7^0\)
\(2x=0\)
\(x=0\)
Ta có abbcca=\(\frac{3}{5}.\frac{4}{5}.\frac{3}{7}\)
=>a2b2c2=\(\frac{36}{175}\)
=>abc=\(\sqrt{\frac{36}{175}}=\frac{6\sqrt{7}}{35}\)
=>a=\(\frac{6\sqrt{7}}{35}:\frac{4}{5}=\frac{3\sqrt{7}}{14}\)=>b=\(\frac{6\sqrt{7}}{35}:\frac{3}{7}=\frac{2\sqrt{7}}{5}\)=>c