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\(\frac{1}{3}x:\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)
\(\Rightarrow\frac{1}{3}x:\frac{2}{3}=\frac{35}{8}\)
\(\Rightarrow\frac{1}{3}x=\frac{35}{12}\)
\(\Rightarrow x=\frac{35}{4}\)
a) \(\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)
\(=\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=\frac{35}{8}\)
\(\Rightarrow\frac{1}{3}\cdot x=\frac{35}{8}\cdot\frac{2}{3}\)
\(\Rightarrow x=\frac{35}{12}:\frac{1}{3}=\frac{35}{4}\)
b) \(4,5:0,3=2,5:\left(0,1\cdot x\right)\)
\(=15=2,5:\left(0,1\cdot x\right)\)
\(\Rightarrow0,1\cdot x=2,25:15\)
\(\Rightarrow x=\frac{3}{20}:0,1=\frac{3}{2}=1,5\)
c) \(8:\left(\frac{1}{4}\cdot x\right)=2:0,02\)
\(8:\left(\frac{1}{4}\cdot x\right)=100\)
\(\Rightarrow\frac{1}{4}\cdot x=8:100\)
\(\Rightarrow x=\frac{2}{25}:\frac{1}{4}=\frac{8}{25}=0,32\)
d) \(3:2\frac{1}{4}=\frac{3}{4}:\left(6\cdot x\right)\)
\(=\frac{4}{3}=\frac{3}{4}:\left(6:x\right)\)
\(\Rightarrow6\cdot x=\frac{3}{4}:\frac{4}{3}\)
\(\Rightarrow x=\frac{9}{16}:6=\frac{3}{32}=0,09375\)
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
làm 1 bài mẫu nha
a) \(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{25}{6}\)
\(\Rightarrow\frac{1}{3}x=\frac{25}{9}\)
\(\Rightarrow x=\frac{25}{9}:\frac{1}{3}\)
\(\Rightarrow x=\frac{25}{3}\)
các bài sau dễ lắm
a) \(\left(\frac{1}{3}x\right):\frac{2}{3}=1\frac{2}{3}:\frac{2}{5}\)
\(\Rightarrow\frac{1}{3}x:\frac{2}{3}=\frac{25}{6}\)
\(\Rightarrow\frac{1}{3}x=\frac{5}{3}\)
\(\Rightarrow x=\frac{5}{3}:\frac{1}{3}\)
\(\Rightarrow x=5\)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
a)\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{7}{4}.\frac{5}{2}\)
\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{35}{8}\)
\(\frac{1}{3}.x=\frac{35}{8}.\frac{2}{3}\)
\(\frac{1}{3}.x=\frac{35}{12}\)
\(x=\frac{35}{12}:\frac{1}{3}\)
\(x=\frac{35}{12}.\frac{3}{1}\)
\(x=\frac{35}{4}\)
b)\(\frac{4,5}{0,3}=\frac{2,25}{0,1.x}\)
\(4,5.\left(0,1.x\right)=2,25.0,3\)
\(4,5.\left(0,1.x\right)=0,675\)
\(0,1.x=0,675:4,5\)
\(0,1.x=0,15\)
\(x=0,15:0,1\)
\(x=1.5\)
c)\(\frac{8}{\frac{1}{4}.x}=\frac{2}{0,02}\)
\(\left(\frac{1}{4}.x\right).2=8.0,02\)
\(\left(\frac{1}{4}.x\right).2=0,16\)
\(\frac{1}{4}.x=0,16:2\)
\(\frac{1}{4}.x=0,08\)
\(x=0,08:\frac{1}{4}\)
\(x=0,32\)
d)\(\frac{3}{\frac{9}{4}}=\frac{\frac{3}{4}}{6.x}\)
\(3.\left(6.x\right)=\frac{3}{4}.\frac{9}{4}\)
\(3.\left(6.x\right)=\frac{27}{16}\)
\(6.x=\frac{27}{16}:3\)
\(6.x=\frac{9}{16}\)
\(x=\frac{9}{16}:6\)
\(x=\frac{3}{32}\)
HỌC TỐT ^^
a) \(\frac{2}{3}=\frac{-10}{x}\)
\(\Rightarrow2x=-30\)
\(\Rightarrow x=-15\)
b) -2|x - 1| = \(\frac{-3}{4}\)
\(\Rightarrow\)|x - 1| = \(\frac{3}{8}\)
\(\Rightarrow\)x - 1 = \(\frac{3}{8}\)hoặc\(\frac{-3}{8}\)
\(\Rightarrow\)x = \(1\frac{3}{8}\)hoặc\(1\frac{-3}{8}\)
a)4,5:0,3=2,25:(0,1.x)
2,25:(0,1.x)=15
0,1.x=0,15
x=0,15:0,1
x=1,5
Vậy x=1,5
b)8:\(\left(\frac{1}{4}.x\right)\)=2:0,02
8:\(\left(\frac{1}{4}.x\right)\)=100
\(\left(\frac{1}{4}.x\right)\)=\(\frac{8}{100}\)
x=\(\frac{8}{100}\):\(\frac{1}{4}\)
x=\(\frac{8}{25}\)
Vậy x=\(\frac{8}{25}\)
c)\(3:2\frac{1}{4}=\frac{3}{4}:\left(6.x\right)\)
\(\frac{3}{4}:\left(6.x\right)=\frac{4}{3}\)
6.x=\(\frac{3}{4}:\frac{4}{3}\)
6.x=\(\frac{9}{16}\)
x=\(\frac{9}{16}\):6
x=\(\frac{3}{32}\)
Vậy x=\(\frac{3}{32}\)
a.
\(4,5\div0,3=2,25\div\left(0,1\times x\right)\)
\(2,25\div\left(0,1\times x\right)=15\)
\(0,1\times x=\frac{2,25}{15}\)
\(0,1\times x=0,15\)
\(x=\frac{0,15}{0,1}\)
\(x=1,5\)
b.
\(8\div\left(\frac{1}{4}\times x\right)=2\div0,02\)
\(8\div\left(\frac{1}{4}\times x\right)=100\)
\(\frac{1}{4}\times x=\frac{8}{100}\)
\(\frac{1}{4}\times x=\frac{2}{25}\)
\(x=\frac{2}{25}\div\frac{1}{4}\)
\(x=\frac{2}{25}\times4\)
\(x=\frac{8}{25}\)
c.
\(3\div2^1_4=\frac{3}{4}\div\left(6\times x\right)\)
\(\frac{3}{4}\div\left(6\times x\right)=3\div2,25\)
\(\frac{3}{4}\div\left(6\times x\right)=\frac{4}{3}\)
\(6\times x=\frac{3}{4}\div\frac{4}{3}\)
\(6\times x=\frac{3}{4}\times\frac{3}{4}\)
\(6\times x=\frac{9}{16}\)
\(x=\frac{9}{16}\div6\)
\(x=\frac{9}{16}\times\frac{1}{6}\)
\(x=\frac{3}{32}\)