Bai 1 : Tim x biet:
a) \(\frac{3}{5}\) cua \(\frac{-14}{15}\) la x
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\(\frac{x-1}{2}\)= \(\frac{2y-4}{6}\)=\(\frac{3z-9}{12}\)=\(\frac{x-1-2y+4+3z-9}{2-6+12}\)= \(\frac{14-1+4-9}{8}\)= 1
=> x =2+1=3
y= (6+4) : 2=5
z=(12+9) : 3=7
Bài 2:
Ta có: \(\left.\begin{matrix} \frac{x}{4} = \frac{y}{5} & & \\ \frac{y}{5} = \frac{z}{2} & & \end{matrix}\right\}\)
=> \(\frac{x}{4} = \frac{y}{5} = \frac{z}{2}\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{4} = \frac{y}{5} = \frac{z}{2} = \frac{x - y + z}{4 - 5 + 2}= \frac{98}{1}= 98\)
=> x = 98 * 4 = 392
y = 98 * 5 = 490
z = 196
Vậy x = 392, y = 490, z = 196
Bài 3:
Gọi x,y lần lượt là số cây trồng của lớp 7A, 7B
Theo đề bài ta có: \(\frac{x}{4} = \frac{y}{5}\) và y - x = 12
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{4} = \frac{y}{5}= \frac{y - x}{5 - 4}= \frac{12}{1}= 12\)
=> x = 12 * 4 = 48
y = 12 * 5= 60
Vậy lớp 7A trồng 48 cây
.......lớp 7B trồng 60 cây
Bài 1: Tìm x, biết
a )24-(36+5)=x b)14-21=(13-x)-(15+8)
24-41=x (13-x)-23=-7
x=-17 13-x=(-7)+23
Vậy x=-17 13-x=16
x=13-16
x=-3 Vậy x=-3
Bài 2:Tìm x, biết
a)17-x=-25+(-16+9) b)3x-21=-19-(-2x)
17-x=-25+(-7) 3x-21=-19+2x
17-x=-32 3x-2x=-19+21
x=17-(-32) x=4
x=49 Vậy x=4
Vậy x=49
Bài 1:
a. 24 - (36+5) = x
=> 24 - 41 = x
=> -17 = x
=> x = -17
b. 14 - 21 = (13 - x) - (15 + 8)
=> -7 = 13 - x - 23
=> -7 - 13 + 23 = -x
=> 3 = -x
=> x = -3
Bài 2:
a. 17 - x = -25 + (-16 + 9)
=> 17 - x = -25 + (-7)
=> 17 - x = -32
=> 17 + 32 = x
=> x = 49
b. 3x - 21 = -19 - (-2x)
=> 3x - 21 = -19 + 2x
=> 3x - 2x = -19 + 21
=> x = 2
\(\frac{x+1}{18}+\frac{x+2}{17}=\frac{x+5}{14}+\frac{x+4}{15}\)
\(\Rightarrow\frac{x+1}{18}+1+\frac{x+2}{17}+1=\frac{x+5}{14}+1+\frac{x+4}{15}+1\)
\(\Rightarrow\frac{x+1}{18}+\frac{18}{18}+\frac{x+2}{17}+\frac{17}{17}=\frac{x+5}{14}+\frac{14}{14}+\frac{x+4}{15}+\frac{15}{15}\)
\(\Rightarrow\frac{x+19}{18}+\frac{x+19}{17}=\frac{x+19}{14}+\frac{x+19}{15}\)
\(\Rightarrow\frac{x+19}{18}+\frac{x+19}{17}-\frac{x+19}{14}-\frac{x+19}{15}=0\)
\(\Rightarrow\left(x+19\right).\left(\frac{1}{18}+\frac{1}{17}-\frac{1}{14}-\frac{1}{15}\right)=0\)
\(\text{Mà }\left(\frac{1}{18}+\frac{1}{17}-\frac{1}{14}-\frac{1}{15}\right)\ne0\text{ nên: }x+19=0\Rightarrow x=-19\)
Ta có : \(\frac{x+2}{198}+\frac{x+3}{197}=\frac{x+4}{196}+\frac{x+5}{195}\)
=> \(\left(\frac{x+2}{198}+1\right)+\left(\frac{x+3}{197}+1\right)=\left(\frac{x+4}{196}+1\right)+\left(\frac{x+5}{195}+1\right)\)
=> \(\frac{x+2+198}{198}+\frac{x+3+197}{197}=\frac{x+4+196}{196}+\frac{x+5+195}{195}\)
=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)
=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)
=> \(\left(x+200\right)\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)=0\)
Ta có : \(\frac{1}{198}+\frac{1}{197}\ne\frac{1}{196}+\frac{1}{195}\) => \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\ne0\)
=> x + 200 = 0
=> x = -200
<=> (\(\frac{x+2}{198}\)+1) +(\(\frac{x+3}{197}\)+1) =(\(\frac{x+4}{196}\)+1) +(\(\frac{x+5}{195}\)+1)
<=> \(\frac{x+200}{198}+\frac{x+200}{197}=\frac{x+200}{196}+\frac{x+200}{195}\)
<=> \(\frac{x+200}{198}+\frac{x+200}{197}-\frac{x+200}{196}-\frac{x+200}{195}=0\)
<=> \(\left(x+200\right)\cdot\left(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\right)\)=0
Vì \(\frac{1}{195}>\frac{1}{196}>\frac{1}{197}>\frac{1}{198}\)
<=> \(\frac{1}{198}+\frac{1}{197}-\frac{1}{196}-\frac{1}{195}\) khác 0
<=> \(x+200=0\)
<=> x =
\(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\)
\(\frac{x}{15}=\frac{-1}{15}\)
=> \(x=-1\)
\(\frac{x}{182}=\frac{-6}{14}\cdot\frac{35}{91}\)
\(\frac{x}{182}=\frac{-15}{91}\)
=> \(91x=182\cdot\left(-15\right)\)
=> \(91x=-2730\)
=> \(x=-30\)
g, \(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\Leftrightarrow\frac{x}{15}=\frac{3}{5}-\frac{2}{3}\Leftrightarrow\frac{x}{15}=-\frac{1}{15}\)
\(\Leftrightarrow x=-1\)
h, \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Leftrightarrow\frac{x}{182}=-\frac{15}{91}\Leftrightarrow\frac{x}{182}=\frac{-30}{182}\)
\(\Leftrightarrow x=-30\)