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dài :vv
a) \(\left|2x-5\right|=4\Leftrightarrow\hept{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Leftrightarrow\hept{\begin{cases}2x=9\\2x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)
b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}\Leftrightarrow\hept{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)
Bài 1 :
a) \(|2x-5|=4\)
\(\Rightarrow\orbr{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Rightarrow}\orbr{\begin{cases}2x=9\\2x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)
b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)
\(\Rightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)
c) \(\left|\frac{-2}{3}\right|+\left|x-\frac{1}{3}\right|=\left|-1\right|-\left|\frac{-1}{3}\right|\)
\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=1-\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=\frac{2}{3}\)
\(\Rightarrow\left|x-\frac{1}{3}\right|=0\Rightarrow x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)
d) \(\left|-\frac{1}{2}\right|-\left|x+\frac{1}{4}\right|=\left|-\frac{3}{4}\right|\)
\(\Rightarrow\frac{1}{2}-\left|x+\frac{1}{4}\right|=\frac{3}{4}\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=-\frac{1}{4}\)
Vì \(\left|x\right|\ge0\Rightarrow\)ko có gtri nào của x thỏa mãn đề bài
Bài 2 :
a) \(\left|x-1\right|=3x+2\)
\(\Rightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}x-3x=2+1\\x+3x=-2+1\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}-2x=3\\4x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{-1}{4}\end{cases}}\)
b|) \(\left|9+x\right|=2x\Rightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-2x=-9\\x+2x=-9\end{cases}\Rightarrow\orbr{\begin{cases}-x=-9\\3x=-9\end{cases}\Rightarrow}\orbr{\begin{cases}x=9\\x=-3\end{cases}}}\)
c) \(\left|x+6\right|-9=2x\Rightarrow\left|x+6\right|=2x+9\)
\(\Rightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=-2x-9\end{cases}\Rightarrow}\orbr{\begin{cases}x-2x=9-6\\x+2x=-9-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=3\\3x=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}}\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^
a,x/2=y/5
<=> 2x/4=y/5=2x+y/4+5=18/9=2
+,x/2=2 => x=4
+, y/5=2 => y=10
g, x/2=y/5
đặt x/2=y/5=k
=> x=2k ; y=5k
ta có 2k.5k=90
k2.10=90
k2=9
=> k=3 k=-3
+, x/2=2=> x=4 x/2=-2 => x=-4
+, y/5=2 => y=10 y/5=-2 => y=-10
CÁC Ý SAU BN LÀM NỐT NHÉ DỄ MÀ
a) Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{x}{2}=\frac{y}{5}=\frac{2x+y}{4+5}=\frac{18}{9}=2\)
\(\Rightarrow x=4;y=10\)
mấy bài còn lại tương tự
a) \(\left(6\frac{2}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)
=> \(\left(\frac{44}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}=-\frac{11}{7}\)
=> \(\frac{44}{7}x+\frac{3}{7}=-\frac{5}{7}\)
=> \(\frac{44}{7}x=-\frac{8}{7}\)
=> \(\frac{44x}{7}=-\frac{8}{7}\)
=> 44x = -8 => 11x = -2 => \(x=-\frac{2}{11}\)
b) \(3\frac{1}{4}x+\left(-\frac{7}{6}\right)-1\frac{2}{3}=\frac{5}{12}\)
=> \(\frac{13}{4}x-\frac{7}{6}-1\frac{2}{3}=\frac{5}{12}\)
=> \(\frac{13}{4}x-\frac{7}{6}=\frac{25}{12}\)
=> \(\frac{13}{4}x=\frac{13}{4}\)
=> x = 1
c) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
=> \(\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)
d) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
=> \(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> \(\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)
e) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{-24}{27}\)
=> \(\left(3x-\frac{7}{9}\right)^3=-1\frac{5}{27}-\left(-\frac{24}{27}\right)=-\frac{32}{27}+\frac{24}{27}=-\frac{8}{27}\)
=> \(\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
=> \(3x-\frac{7}{9}=-\frac{2}{3}\)
=> \(x=\frac{-\frac{2}{3}+\frac{7}{9}}{3}=\frac{1}{27}\)
g) \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{99\cdot100}=\frac{99}{100}\)
=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{99}-\frac{x}{100}=\frac{99}{100}\)
=> \(\frac{x}{1}-\frac{x}{100}=\frac{99}{100}\)
=> \(\frac{100x-x}{100}=\frac{99}{100}\)
=> \(\frac{99x}{100}=\frac{99}{100}\)
=> x = 1
h) \(\frac{x}{3}+\frac{x}{6}+\frac{x}{10}+\frac{x}{15}=3x-1\)
=> \(\frac{2x}{6}+\frac{2x}{12}+\frac{2x}{20}+\frac{2x}{30}=3x-1\)
=> \(\frac{2x}{2\cdot3}+\frac{2x}{3\cdot4}+\frac{2x}{4\cdot5}+\frac{2x}{5\cdot6}=3x-1\)
=> \(2\left(\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}\right)=3x-1\)
=> \(2\left(\frac{x}{2}-\frac{x}{6}\right)=3x-1\)
=> \(2\left(\frac{3x}{6}-\frac{x}{6}\right)=3x-1\)
=> \(2\cdot\frac{2x}{6}=3x-1\)
=> \(\frac{x}{3}=\frac{3x-1}{2}\)
=> 2x = 3(3x - 1)
=> 2x - 9x + 3 = 0
=> -7x = -3
=> x = 3/7
a. Vì \(\left|x-y-5\right|\ge0\forall x;y;2019\left|y-3\right|^{2020}\ge0\forall y\)
\(\Rightarrow\left|x-y-5\right|+2019\left|y-3\right|^{2020}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-y-5\right|=0\\2019\left|y-3\right|^{2020}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y-5=0\\y-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=5\\y=3\end{cases}}\)
b. \(2\left(x-5\right)^4\ge0\forall x;5\left|2y-7\right|^5\ge0\forall y\)
\(\Rightarrow2\left(x-5\right)^4+5\left|2y-7\right|^5\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}2\left(x-5\right)^4=0\\5\left|2y-7\right|^5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-5=0\\2y-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\y=\frac{7}{2}\end{cases}}\)
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
\(b,\frac{z}{7}=-\frac{11}{-28}\)
\(\Leftrightarrow z.\left(-28\right)=-11.7\)
\(\Leftrightarrow z.\left(-28\right)=-77\)
\(\Leftrightarrow z=\frac{11}{4}\)
\(a,-\frac{2}{3}=\frac{x-3}{-6}=\frac{10}{5-y}=\frac{4-2z}{9}\)
Xét :
\(-\frac{2}{3}=\frac{x-3}{-6}\)
\(\Leftrightarrow-2.\left(-6\right)=\left(x-3\right).3\)
\(\Leftrightarrow12=\left(x-3\right).3\)
\(\Leftrightarrow4=x-3\Leftrightarrow x=7\)
Xét
\(-\frac{2}{3}=\frac{10}{5-y}\)
\(\Leftrightarrow-2.\left(5-y\right)=10.3\)
\(\Leftrightarrow-10+2y=30\)
\(\Leftrightarrow2y=40\Leftrightarrow y=20\)
Xét :
\(-\frac{2}{3}=\frac{4-2z}{9}\)
\(\Leftrightarrow-2.9=\left(4-2z\right).3\)
\(\Leftrightarrow-18=\left(4-2z\right).3\)
\(\Leftrightarrow-6=4-2z\)
\(\Leftrightarrow10=2z\Leftrightarrow z=5\)
Vậy \(\left(x;y;z\right)=\left(7;20;5\right)\)
a) theo công thức ta có: 5.7=x.-y=5.7
\(\Rightarrow x=-7;y=-5\)
b) \(\frac{x}{4}=\frac{y}{3}=\frac{x+y}{4+3}=\frac{14}{7}=2\)
\(\Rightarrow\frac{x}{4}=2\Rightarrow x=2.4=8\)
\(\Leftrightarrow\frac{y}{3}=2\Rightarrow y=2.3=6\)
Vậy x=8; y=6
Bạn giải thích giùm mình tại sao\(\frac{x+y}{4+3}\)\(=\)\(\frac{14}{7}\)\(=\)\(2\)