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Bài 1:
Ta có: \(3x=2y\)
nên \(\dfrac{x}{2}=\dfrac{y}{3}\)
mà x+y=-15
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)
Vậy: (x,y)=(-6;-9)
Bài 2:
a) Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}\)
mà x+y-z=20
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{4+3-5}=\dfrac{20}{2}=10\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{4}=10\\\dfrac{y}{3}=10\\\dfrac{z}{5}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=40\\y=30\\z=50\end{matrix}\right.\)
Vậy: (x,y,z)=(40;30;50)
a) \(-\dfrac{3}{5}-x=-0,75\)
\(\Rightarrow-\dfrac{3}{5}-x=-\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{15}{20}-\dfrac{12}{20}=\dfrac{8}{20}=\dfrac{2}{5}\)
b) \(1\dfrac{4}{5}=-0,15-x\)
\(\Rightarrow\dfrac{9}{5}=-\dfrac{3}{20}-x\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{9}{5}\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{36}{20}=-\dfrac{39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{2}{5}\)
a) \(-\dfrac{3}{5}-x=-0,75\)
\(x=-\dfrac{3}{5}+0,75=\dfrac{3}{5}+\dfrac{3}{4}\)
\(x=\dfrac{27}{20}\)
________
b) \(1\dfrac{4}{5}=-0,15-x\)
\(=>-0,15-x=\dfrac{9}{5}\)
\(x=\dfrac{-3}{20}-\dfrac{9}{5}=\dfrac{-3}{20}-\dfrac{36}{20}\)
\(x=\dfrac{-39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)=\dfrac{6}{15}+\dfrac{5}{15}\)
\(x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(x=\dfrac{11}{15}-\dfrac{1}{3}=\dfrac{11}{15}-\dfrac{5}{15}\)
\(x=\dfrac{6}{15}=\dfrac{2}{5}\)
\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)
\(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
a. \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\dfrac{1}{3}\right|=-1+4=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\)
Vậy..........
b. \(1\dfrac{3}{4}.x+1\dfrac{1}{2}=-\dfrac{4}{5}\)
\(\Rightarrow1\dfrac{3}{4}x=-\dfrac{4}{5}-1\dfrac{1}{2}=\dfrac{-23}{10}\)
\(\Rightarrow x=\dfrac{-23}{10}:1\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{-46}{35}\)
\(a,x-\dfrac{2}{3}=\dfrac{1}{6}\\ \Rightarrow x=\dfrac{5}{6}\\ b,2x+\dfrac{1}{2}=-\dfrac{5}{3}\\ \Rightarrow2x=-\dfrac{13}{6}\\ \Rightarrow x=-\dfrac{13}{12}\\ c,3x+\dfrac{3}{2}=x-\dfrac{5}{3}\\ \Rightarrow-4x+\dfrac{3}{2}=-\dfrac{5}{3}\\ \Rightarrow-4x=-\dfrac{19}{6}\\ \Rightarrow4x=\dfrac{19}{6}\\ \Rightarrow x=\dfrac{19}{24}.\)
a) x - 2/3 = 1/6
x = 1/6 + 2/3
x = 5/6
b) 2x + 1/2 = -5/3
2x = -5/3 - 1/2
2x = -13/6
x = -13/6 : 2
x = -13/12
c) 3x + 3/2 = x - 5/3
3x - x = -5/3 - 3/2
2x = -19/6
x = -19/6 : 2
x = -19/12
a) x = 1
⇒ 3a - 1 = 5
⇒ 3a = 6
⇒ a = 2
b) x = 5
⇒ 3a - 1 = 1
⇒ 3a = 2
⇒ a = 2/3
a) \(\dfrac{5}{3a-1}=1\)
\(\Rightarrow3a-1=5\)
\(\Rightarrow3a=6\)
\(\Rightarrow a=\dfrac{6}{3}=2\)
b) \(\dfrac{5}{3a-1}=-5\)
\(\Rightarrow3a-1=5:\left(-5\right)=-1\)
\(\Rightarrow3a=-1+1=0\)
\(\Rightarrow a=0:3=0\)
a: \(\left|a-2b+3\right|^{2023}>=0\forall a,b\)
\(\left(b-1\right)^{2024}>=0\forall b\)
Do đó: \(\left|a-2b+3\right|^{2023}+\left(b-1\right)^{2024}>=0\forall a,b\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}a-2b+3=0\\b-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=1\\a=2b-3=2\cdot1-3=-1\end{matrix}\right.\)
Thay a=-1 và b=1 vào P, ta được:
\(P=\left(-1\right)^{2023}\cdot1^{2024}+2024=2024-1=2023\)
\(a,1\dfrac{1}{5}+\dfrac{4}{5}:x=0,75\\ \dfrac{ 4}{5}:x=\dfrac{3}{4}-\dfrac{6}{4}\\ \dfrac{4}{5}:x=-\dfrac{3}{4}\\ x=\dfrac{4}{5}:\left(-\dfrac{3}{4}\right)\\ x=-\dfrac{16}{15}\\ b,x+\dfrac{1}{2}=1-x\\ x+x=1-\dfrac{1}{2}\\ 2x=\dfrac{1}{2}\\ x=\dfrac{1}{2}:2\\ x=\dfrac{1}{4}\)
$ \dfrac{1}{5}+\dfrac{4}{5}: x=0,75$;
b) $x+\dfrac{1}{2}=1-x$.
Em bị trục trặc và nhìn thấy phần câu hỏi bị thế này ạ. Em nhìn em không giải được ạ