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a, 1+2y / 18 = 1+4y / 24 = 1+6y / 6x
Ta có : 1+2y / 18 = 1+6y / 6x = 1+2y + 1+6y / 18 + 6y
= 2+ 8y / 18+6y = 2 (1+4y) / 2( 9 +3y) = 1+4y/9+3y
Ta lại có : 1 + 4y/24 = 1+4y / 9+3y
=> 24=9+3y => 15=3y => y=5
Vậy y=5
Nhớ like
b, 1+3y/12 = 1+5y/5x = 1+7y/4x
Ta có : 1+3y/12 = 1+7y/4x = 1+3y+1+7y / 12 +4x
= 2 + 10y / 12 +4x = 2 (1+5y) / 2 (6+2x) = 1+5y / 6+2x
Ta lại có: 1+5y / 5x = 1+5y / 6+2x
=> 5x = 6+2x => 3x = 6 => x=2
Vậy x =2
a)\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1980}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1980}{1975}\\z=2004\end{matrix}\right.\)
b) \(\left|\dfrac{3}{4}+x\right|+\left|-\dfrac{1}{5}+y\right|+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{4}+x\right|=0\\\left|-\dfrac{1}{5}+y\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)
1.
\(\left(\dfrac{-2}{3}\right).0,75+1\dfrac{2}{3}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)^2\)
\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(\dfrac{9}{-4}\right)+\dfrac{1}{4}\)
\(=-\dfrac{1}{2}+\dfrac{45}{-12}+\dfrac{1}{4}\)
\(=-\dfrac{6}{12}+\dfrac{-45}{12}+\dfrac{3}{4}\)
\(=\dfrac{-48}{12}\)
\(=-4\)
2.
a) \(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{10}{20}\)
\(\Leftrightarrow x=\dfrac{-11}{20}\)
b) \(\left|x-\dfrac{2}{5}\right|+\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{11}{4}-\dfrac{3}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=-2\Rightarrow x=-2+\dfrac{2}{5}=\dfrac{-8}{5}\\x-\dfrac{2}{5}=2\Rightarrow x=2+\dfrac{2}{5}=\dfrac{12}{5}\end{matrix}\right.\)
3.
a) \(\dfrac{16}{2^n}=2\)
\(\Leftrightarrow2^n=16:2\)
\(\Leftrightarrow2^n=8\)
\(\Leftrightarrow2^n=2^3\)
\(\Leftrightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^n=\left(-27\right).81\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)
\(\Leftrightarrow n=7\)
4. Ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)
\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)
Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Vì \(x-y+x=-49\) ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)
Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)
mình nghĩ bạn chép sai đề bài
dấu ''='' thứ 2 thay bằng dấu ''+''
ta có
\(\dfrac{19}{x+y}+\dfrac{19}{y+z}+\dfrac{19}{x+z}=\dfrac{133}{10}\)
\(\Rightarrow19\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{133}{10}\)
\(\Rightarrow\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}=\dfrac{7}{10}\)
lại có
\(\dfrac{7x}{y+z}+\dfrac{7y}{x+z}+\dfrac{7z}{x+y}=\dfrac{133}{10}\)
\(\Rightarrow7\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)=\dfrac{133}{10}\)
\(\Rightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{19}{10}\)
\(\Rightarrow\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}+\dfrac{x+y+z}{x+y}=\dfrac{49}{10}\)
\(\Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)=\dfrac{49}{10}\)
\(\Rightarrow\dfrac{7}{10}\left(x+y+z\right)=\dfrac{49}{10}\Rightarrow\left(x+y+z\right)^2=49.\)
\(\dfrac{x-2}{2}=\dfrac{y-4}{3}=\dfrac{z-8}{5}\)
\(\Rightarrow\dfrac{x-2}{2}+2=\dfrac{y-4}{3}+2=\dfrac{z-8}{5}+2\)
\(\Rightarrow\dfrac{x+2}{2}=\dfrac{y+2}{3}=\dfrac{z+2}{5}\)
\(\Rightarrow\left(\dfrac{x+2}{2}\right)^2=\left(\dfrac{y+2}{3}\right)^2=\left(\dfrac{z+2}{5}\right)^2\)
\(\Rightarrow\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{\left(x+2\right)^2}{4}=\dfrac{\left(y+2\right)^2}{9}=\dfrac{\left(z+2\right)^2}{25}=\dfrac{3.\left(y+2\right)^2}{27}\dfrac{\left(x+2\right)^2+3\left(y+2\right)^2-\left(z+2\right)^2}{4+27-25}=\dfrac{24}{6}=4\)\(\Rightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=16\\\left(y+2\right)^2=36\\\left(z+2\right)^2=100\end{matrix}\right.\)
Bạn chia trường hợp rồi tìm x,y,z nhé
1) Phân số đầu nhân 2.
_ Phân số thứ 2 nhân 3, p/s thứ 3 giữ nguyên.
_ Lấy phân số đầu + p/s thứ 2 - p/s thứ 3.
_ Dựa vào dãy tỉ số bằng nhau tìm x, y, z.
2) \(x-y-z=0\Rightarrow x=y+z\)
Khi đó thay vào B được:
\(B=\left(1-\dfrac{z}{y+z}\right)\left(1-\dfrac{y+z}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{y}{y+z}.\dfrac{z}{y}.\dfrac{y+z}{z}\)
\(=1\)
Vậy B = 1.
Ta có :
\(\dfrac{x+y-z}{z}=\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}\\ \Leftrightarrow\dfrac{x+y+z}{z}=\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\left(cùngcộngthêm2\right)\)
TH1: \(x+y+z\ne0\)
\(\Rightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)\\ =2\cdot2\cdot2=8\)
TH2: \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)(*)
\(\Rightarrow P=\left(1+\dfrac{-\left(y+z\right)}{y}\right)\left(1+\dfrac{-\left(z+x\right)}{z}\right)\left(1+\dfrac{-\left(x+y\right)}{z}\right)\\ =\left(1-1-\dfrac{z}{y}\right)\left(1-1-\dfrac{x}{z}\right)\left(1-1-\dfrac{y}{z}\right)\\ =\left(-\dfrac{z}{y}\right)\left(-\dfrac{x}{z}\right)\left(-\dfrac{y}{z}\right)\\ =-1\)
Vậy P=8 hoặc P=-1
\(\dfrac{\left|x\right|}{11}=\dfrac{y}{12};7y=3z\)
\(\Rightarrow\dfrac{\left|x\right|}{11}=\dfrac{y}{12};\dfrac{y}{3}=\dfrac{z}{7}\)
\(\Rightarrow\dfrac{\left|x\right|}{11}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{\left|x\right|}{11}=\dfrac{y}{12}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{2\left|x\right|}{22}=\dfrac{y}{12}=\dfrac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2\left|x\right|}{22}=\dfrac{y}{12}=\dfrac{z}{28}\)
\(=\dfrac{2\left|x\right|+y+z}{22+12+28}\)
\(=\dfrac{152}{62}=\dfrac{76}{31}\)
Áp dụng tính x;y;z