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a ) \(\left(-\frac{40}{51}.0,32.\frac{17}{20}\right):\frac{64}{75}\)
\(=\left(-\frac{40}{51}.\frac{8}{25}.\frac{17}{20}\right):\frac{64}{75}\)
\(=\left(\frac{-40.8.17}{51.25.20}\right):\frac{64}{75}\)
\(=\left(\frac{-16}{75}\right).\frac{75}{64}\)
\(=\frac{-1}{1}.\frac{1}{4}=-\frac{1}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-9=15k\\y-12=20k\\z-24=40k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=15k+9\\y=20k+12\\z=40k+24\end{array}\right.}\)
ta có:
x.y=1200\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\Rightarrow\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}=k\)
=> (15k+9)(20k+12)=1200
=> 3.4(5k+3)(5k+3)=1200
=> (5k+3)2=100
=> 5k+3=\(\pm\)10
=> \(\left[\begin{array}{nghiempt}5k+3=10\\5k+3=-10\end{cases}\Rightarrow\left[\begin{array}{nghiempt}5k=7\\5k=-13\end{cases}\Rightarrow}\left[\begin{array}{nghiempt}k=\frac{7}{5}\\k=-\frac{13}{5}\end{array}\right.}\)
* với k=7/5
x=7/5x15+9=30
y=7/5x20+12=40
z=7/5x40+24=80
* với k=-13/5
x=-13/5x15+9=-30
y=-13/5x20+12=-40
z=-13/5x40+24=-80
b)
\(\frac{40}{x-30}=\frac{20}{y-50}=\frac{28}{z-21}\Rightarrow\frac{x-30}{40}=\frac{y-50}{20}=\frac{z-21}{28}k=\)
=>\(\left[\begin{array}{nghiempt}x-30=40k\\y-50=20k\\z-21=28k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=40k+30\\y=20k+50\\z=28k+21\end{array}\right.}\)
ta có:
x.y.z=22400
=> (40k+30)(20k+50)(28k+21)=22400
c) 15x=-10y=6z
\(\Rightarrow\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\Rightarrow\frac{x}{2}=-\frac{y}{3}=\frac{z}{5}=k\)
=> \(\left[\begin{array}{nghiempt}x=2k\\y=-3k\\z=5k\end{array}\right.\)
ta có:
x.y.z=30000
=> 2k.(-3k).5k=30000
=> k3=1000
=> k=10
ta có: x=10x2=20
y=10.(-3)=-30
z=10.5=50
\(\frac{1}{20}\left(x-\frac{8}{15}\right)=-\frac{1}{30}\) \(\left(28+\frac{1}{5}\right).\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{1}{30}:\frac{1}{20}\) \(\frac{141}{5}.\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{2}{3}\) \(\frac{3}{5}.x+\frac{4}{7}=0\)
\(x=-\frac{2}{3}+\frac{8}{15}\) \(\frac{3}{5}.x=-\frac{4}{7}\)
\(x=-\frac{2}{15}\) \(x=-\frac{20}{21}\)
cộng thêm 1 của mỗi đẳng thức :
\(\frac{a}{b+c}+1=\frac{c}{a+b}+1=\frac{b}{c+a}+1\)
hay \(\frac{a+b+c}{b+c}=\frac{a+b+c}{a+b}=\frac{a+b+c}{c+a}\)
với a + b + c = 0 thì :
b + c = -a ; a + b = -c ; c + a = -b
nên \(20.\left(\frac{a}{b+c}\right)+3.\left(\frac{c}{a+b}\right)+1998.\left(\frac{b}{c+a}\right)=20.\left(\frac{a}{-a}\right)+3.\left(\frac{c}{-c}\right)+1998.\left(\frac{b}{-b}\right)\)
hay \(20.\left(-1\right)+3.\left(-1\right)+1998.\left(-1\right)=-20+\left(-3\right)+\left(-1998\right)=-2021\)
với a + b + c khác 0 thì : a = b = c
nên \(20.\left(\frac{a}{b+c}\right)+3.\left(\frac{c}{a+b}\right)+1998.\left(\frac{b}{c+a}\right)=20.\frac{1}{2}+3.\frac{1}{2}+1998.\frac{1}{2}=\frac{2021}{2}\)
Nếu a+b+c = 0 => Biểu thức = 20.(-1)+3.(-1)+1998.(-1) = -2021
Nếu a+b+c khác 0 thì :
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
a/b+c = c/a+b = b/c+a = a+b+c/2a+2b+2c = 1/2
=> Biểu thức = 20.1/2+3.1/2+1998.1/2 = 2021/2
Vậy ............
k mk nha
Lời giải:
$\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}$
$=\frac{5(3a-2b)}{25}=\frac{3(2c-5a)}{9}=\frac{2(5b-3c)}{4}$
$=\frac{5(3a-2b)+3(2c-5a)+2(5b-3c)}{25+9+4}=\frac{0}{25+9+4}=0$
$\Rightarrow 3a-2b=2c-5a=5b-3c=0$
$\Rightarrow 3a=2b; 2c=5a$
$\Rightarrow \frac{a}{2}=\frac{b}{3}=\frac{c}{5}$
Áp dụng TCDTSBN:
$\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{2+3+5}=\frac{-50}{10}=-5$
$\Rightarrow a=(-5).2=-10; b=(-5).3=-15; c=(-5).5=-25$
Gọi \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)(1)
Thay (1) vào ta có :
\(\frac{5a+3b}{5a-3b}=\frac{5kb+3b}{5kb-3b}=\frac{b\left(5k-3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\)(2)
\(\frac{5c+3d}{5c-3d}=\frac{5kd+3d}{5kd-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\)(3)
Từ (2) và (3)
\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
\(\RightarrowĐPCM\)
\(\Leftrightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(2c+13d\right)\left(3a-7b\right)\)
\(\Leftrightarrow6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd\)
\(\Leftrightarrow-14ad+14bc=39ad-39bc\)
\(\Leftrightarrow-14\left(ad-bc\right)=39\left(ad-bc\right)\)
=>ad-bc=0
=>ad=bc
hay a/b=c/d
Ta có: \(\frac{a}{a'}+\frac{b}{b'}=1\)
\(\Rightarrow\frac{a}{a'}.\frac{b}{b'}+\frac{b'}{b}.\frac{b}{b'}=\frac{b}{b'}.\)
\(\Rightarrow\frac{ab}{a'b'}+1=\frac{b}{b'}\) (1).
\(\frac{b}{b'}+\frac{c'}{c}=1\)
\(\Rightarrow\frac{b}{b'}=1-\frac{c'}{c}\) (2).
Từ (1) và (2) => \(\frac{ab}{a'b'}=-\frac{c'}{c}\)
\(\Rightarrow abc=-a'b'c'\)
\(\Rightarrow abc+a'b'c'=0\left(đpcm\right).\)
Vậy \(abc+a'b'c'=0.\)
Chúc bạn học tốt!