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Bài 10:
1: \(\left(7-\frac15+\frac13\right)-\left(6+\frac95+\frac43\right)\)
\(=7-\frac15+\frac13-6-\frac95-\frac43\)
\(=\left(7-6\right)+\left(-\frac15-\frac95\right)+\left(\frac13-\frac43\right)\)
=1-2-1
=-2
2: \(7+\left(\frac{7}{12}-\frac12+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)
=10-5
=5
3: \(\left(\frac12-\frac13\right)-\left(\frac53-\frac32\right)+\left(\frac73-\frac52\right)\)
\(=\frac12-\frac13-\frac53+\frac32+\frac73-\frac52\)
\(=-\frac12+\frac13=\frac{-3+2}{6}=-\frac16\)
4: \(\left(\frac27-\frac94\right)-\left(-\frac37+\frac54\right)-\left(\frac24-\frac97\right)\)
\(=\frac27-\frac94+\frac37-\frac54-\frac24+\frac97\)
\(=\left(\frac27+\frac37+\frac97\right)+\left(-\frac94-\frac54-\frac24\right)=\frac{14}{7}-\frac{16}{4}=2-4=-2\)
5: \(\left(\frac53-\frac37+9\right)-\left(2+\frac57-\frac23\right)+\left(\frac87-\frac43-10\right)\)
\(=\frac53-\frac37+9-2-\frac57+\frac23+\frac87-\frac43-10\)
\(=\left(\frac53+\frac23-\frac43\right)+\left(-\frac37-\frac57+\frac87\right)+\left(9-2-10\right)\)
\(=\frac33+\left(-3\right)=1-3=-2\)
Bài 11:
1: \(\frac25\cdot\frac38_{}+\frac58\cdot\frac25=\frac25\left(\frac38+\frac58\right)=\frac25\cdot\frac88=\frac25\)
2: \(\frac23\cdot\frac52-\frac34\cdot\frac23=\frac23\left(\frac52-\frac34\right)=\frac23\cdot\frac74=\frac{14}{12}=\frac76\)
3: \(\frac57\cdot\frac{19}{23}-\frac{12}{23}\cdot\frac57=\frac57\left(\frac{19}{23}-\frac{12}{23}\right)=\frac57\cdot\frac{7}{23}=\frac{5}{23}\)
4: \(\frac72\cdot\frac{11}{6}-\frac72\cdot\frac56=\frac72\left(\frac{11}{6}-\frac56\right)=\frac72\cdot\frac66=\frac72\)
5: \(\frac{11}{9}\cdot\frac34-\frac29\cdot\frac34=\frac34\left(\frac{11}{9}-\frac29\right)=\frac34\cdot\frac99=\frac34\)
6: \(\frac37\cdot\frac{13}{5}+\frac37\cdot\frac85=\frac37\left(\frac{13}{5}+\frac85\right)=\frac37\cdot\frac{21}{5}=\frac{21}{7}\cdot\frac35=3\cdot\frac35=\frac95\)
7: \(\frac{7}{15}\cdot\frac{16}{13}+\frac{7}{15}\cdot\frac{-3}{13}=\frac{7}{15}\left(\frac{16}{13}-\frac{3}{13}\right)=\frac{7}{15}\cdot\frac{13}{13}=\frac{7}{15}\)
8: \(-\frac{23}{7}\cdot\frac{3}{10}+\frac{13}{7}\cdot\frac{3}{10}=\frac{3}{10}\left(-\frac{23}{7}+\frac{13}{7}\right)=\frac{3}{10}\cdot\frac{-10}{7}=-\frac37\)
9: \(\frac{-11}{8}\cdot\frac{19}{3}+\frac{19}{3}\cdot\frac{-5}{8}=\frac{19}{3}\left(-\frac{11}{8}-\frac58\right)=\frac{19}{3}\cdot\left(-2\right)=-\frac{38}{3}\)
Bài 12: Bài 12:
1: \(\frac{-5}{17}\cdot\frac{31}{33}+\frac{-5}{17}\cdot\frac{2}{33}+1\frac{5}{17}\)
\(=-\frac{5}{17}\cdot\left(\frac{31}{33}+\frac{2}{33}\right)+1+\frac{5}{17}\)
\(=-\frac{5}{17}+1+\frac{5}{17}=1\)
2: \(\frac57\cdot\left(-\frac{3}{11}\right)+\frac57\cdot\left(-\frac{8}{11}\right)+2\frac57\)
\(=-\frac57\left(\frac{3}{11}+\frac{8}{11}\right)+2+\frac57\)
\(=-\frac57+2+\frac57=2\)
3: \(\frac{9}{10}\cdot\frac{23}{11}-\frac{1}{11}\cdot\frac{9}{10}+\frac{9}{10}\)
\(=\frac{9}{10}\left(\frac{23}{11}-\frac{1}{11}+1\right)\)
\(=\frac{9}{10}\cdot\left(2+1\right)=\frac{9}{10}\cdot3=\frac{27}{10}\)
4: \(\frac54\cdot\frac{8}{15}+\frac{-5}{16}\cdot\frac{8}{15}-1\)
\(=\frac{8}{15}\left(\frac54-\frac{5}{16}\right)-1\)
\(=\frac{8}{15}\left(\frac{20}{16}-\frac{5}{16}\right)-1=\frac{8}{16}-1=-\frac{8}{16}=-\frac12\)
5: \(-\frac{19}{3}\cdot\frac{14}{4}+\frac{25}{4}\cdot\frac{-19}{3}+4\frac34\)
\(=-\frac{19}{4}\left(\frac{14}{3}+\frac{25}{3}\right)+4\frac34\)
\(=-\frac{19}{4}\cdot13+\frac{19}{4}=\frac{19}{4}\left(-13+1\right)=\frac{19}{4}\cdot\left(-12\right)=-57\)
6: \(\frac{1}{27}\cdot\frac{-3}{7}-\frac59\cdot\frac{-3}{7}+\frac19\)
\(=-\frac37\left(\frac{1}{27}-\frac59\right)+\frac19\)
\(=-\frac37\left(\frac{1}{27}-\frac{15}{27}\right)+\frac19=-\frac37\cdot\frac{-14}{27}+\frac19=\frac29+\frac19=\frac39=\frac13\) b
Ta có: \(\left(3x-2\right)^{2024}\ge0\forall x\)
=>\(4\left(3x-2\right)^{2024}\ge0\forall x\)
mà \(\left(y+1\right)^{10}\ge0\forall y\)
nên \(4\left(3x-2\right)^{2024}+\left(y+1\right)^{10}\ge0\forall x,y\)
=>\(4\left(3x-2\right)^{2024}+\left(y+1\right)^{10}+2025\ge2025\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}3x-2=0\\ y+1=0\end{cases}\Rightarrow\begin{cases}x=\frac23\\ y=-1\end{cases}\)
Ta có 2 TH:
+ Th1: \(x-2=x\)
=>\(x-x=2\)
=>\(0=2\)( Vô lý, loại)
+ Th2: \(x-2=-x\)
=>\(x+x=2\)
=>\(2x=2\)
=>\(x=1\)
Vậy x=1
\(|x-2|=x\)
\(\Rightarrow TH1:x-2=x\)
\(x-x=2\)
\(0=2\)
\(\Rightarrow x\in\varnothing\)
\(TH2:x-2=-x\)
\(x+x=2\)
\(2x=2\)
\(\Rightarrow x=1\)
Vậy \(x\in\left\{\varnothing;1\right\}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x}{y+z-5}=\frac{y}{x+z+3}=\frac{z}{x+y+2}=\frac{x+y+z}{y+z-5+x+z+3+x+y+2}=\frac{x+y+z}{2x+2y+2z}=\frac12\)
=>\(\begin{cases}y+z-5=2x\\ x+z+3=2y\\ x+y+2=2z\end{cases}\Rightarrow\begin{cases}y+z=2x+5\\ y+z=2y-3\\ x+y=2z-2\end{cases}\)
\(\frac{x}{y+z-5}=\frac12\left(x+y+z\right)\)
=>\(\frac12\left(x+y+z\right)=\frac12\)
=>x+y+z=1
*Ta có: x+y+z=1
=>z+2z-2=1
=>3z-2=1
=>3z=3
=>z=1
*Ta có: x+y+z=1
=>y+2y-3=1
=>3y=4
=>\(y=\frac43\)
*Ta có: x+y+z=1
=>x+2x+5=1
=>3x+5=1
=>3x=-4
=>\(x=-\frac43\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x}{y+z-5}=\frac{y}{x+z+3}=\frac{z}{x+y+2}=\frac{x+y+z}{y+z-5+x+z+3+x+y+2}=\frac{x+y+z}{2x+2y+2z}=\frac12\)
=>\(\begin{cases}y+z-5=2x\\ x+z+3=2y\\ x+y+2=2z\end{cases}\Rightarrow\begin{cases}y+z=2x+5\\ y+z=2y-3\\ x+y=2z-2\end{cases}\)
\(\frac{x}{y+z-5}=\frac12\left(x+y+z\right)\)
=>\(\frac12\left(x+y+z\right)=\frac12\)
=>x+y+z=1
*Ta có: x+y+z=1
=>z+2z-2=1
=>3z-2=1
=>3z=3
=>z=1
*Ta có: x+y+z=1
=>y+2y-3=1
=>3y=4
=>\(y=\frac43\)
*Ta có: x+y+z=1
=>x+2x+5=1
=>3x+5=1
=>3x=-4
=>\(x=-\frac43\)
Bạn ơi mình hiểu rồi , làm cho nó cùng mũ rồi bỏ mũ luôn đúng không ^^
Cám ơn bạn nhiều lắm ^^
Chúc bạn học tốt
Ta có:
\(\frac{x}{2}-\frac{3}{y}=\frac{5}{4}\)
hay \(\frac{2x}{4}-\frac{3}{y}=\frac{5}{4}\)
Suy ra \(\frac{3}{y}=\frac{2x-5}{4}\)
\(\Rightarrow3\cdot4=\left(2x-5\right)y\)
hay \(\left(2x-5\right)y=12\)
Đến đây bạn tự lập bảng giá trị nhé!

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