Tìm x trong tỉ lệ thức:
\(\dfrac{5x+3}{2\dfrac{1}{7}}=\dfrac{\dfrac{7}{15}}{5x+3}\)
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a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
a: \(\Leftrightarrow-2x=-\dfrac{1}{3}-\dfrac{1}{8}-\dfrac{5}{7}=-\dfrac{197}{168}\)
hay x=197/336
c: \(\Leftrightarrow5x=9+\dfrac{6}{18}-\dfrac{2}{7}=\dfrac{190}{21}\)
hay x=38/21
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
a: \(\dfrac{15}{21}=\dfrac{5}{7}=\dfrac{30}{42}\)
=>Lập được TLT
b: \(\dfrac{0.25}{1.25}=\dfrac{1}{5}< >\dfrac{1}{7}\)
=>KO lập được TLT
c: \(0.4:\left(1+\dfrac{2}{5}\right)=0.4:1.4=\dfrac{2}{7}< >\dfrac{3}{5}\)
=>Ko lập được TLT
d: \(\dfrac{3}{5}:\dfrac{1}{7}=\dfrac{21}{5}=< >21:\dfrac{1}{5}\)
=>Ko lập được TLT
e: \(4+\dfrac{1}{2}:7+\dfrac{1}{2}=4.5:7.5=\dfrac{3}{5}< >\dfrac{2.7}{4.7}\)
=>Ko lập được TLT
f: 1/4:1/9=9/4
1/2:2/9=9/4
=>1/4:1/9=1/2:2/9
=>Lập được TLT
g: 2/7:4/11=2/7*11/4=22/28=11/14
7/2:4/11=7/2*11/4=77/8<>11/14
=>Ko lập được TLT
h: 2/5:10/2=2/5*2/10=4/50=2/25
2/1:1/4=8<>2/25
=>Ko lập được TLT
i: 2/7:7/4=2/7*4/7=8/49
16/49:2=8/49=2/7:7/4
=>Lập được TLT
1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(3x+9+4x-12=3x-7\)
\(\Leftrightarrow4x=-7+12-9=-4\)
hay \(x=-1\left(nhận\right)\)
2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(3x+12-4x+16=3x-4\)
\(\Leftrightarrow28-4x=-4\)
\(\Leftrightarrow4x=32\)
hay \(x=8\left(tm\right)\)
3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
Suy ra: \(5x^2-12+3x+3=5x^2-5x\)
\(\Leftrightarrow3x-9+5x=0\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(nhận\right)\)
\(7 : 21 = \dfrac{7}{{21}} = \dfrac{1}{3}\);
\(\dfrac{1}{5}:\dfrac{1}{2} = \dfrac{1}{5} .\dfrac{2}{1} = \dfrac{2}{5}\);
\(\dfrac{1}{4}:\dfrac{3}{4} = \dfrac{1}{4}.\dfrac{4}{3} = \dfrac{1}{3}\);
\( 1,1 : 3,2 = \dfrac{{1,1}}{{3,2}}=\dfrac{11}{32}\);
\(1 : 2,5 =\dfrac{1}{{2,5}}=\dfrac{10}{25}=\dfrac{2}{5}\).
Ta thấy có các tỉ số bằng nhau là :
+) \(\dfrac{1}{4}:\dfrac{3}{4}\) và \(7 : 21\) (vì cùng bằng \(\dfrac{1}{3}\)) nên ta có tỉ lệ thức : \(\dfrac{1}{4}:\dfrac{3}{4} = 7:21\).
+) \(\dfrac{1}{5}:\dfrac{1}{2}\) và \(1 : 2,5\) (vì cùng bằng \(\dfrac{2}{5}\)) nên ta có tỉ lệ thức : \(\dfrac{1}{5}:\dfrac{1}{2} = 1 : 2,5\).
Lời giải:
\(\frac{5x+3}{2\frac{1}{7}}=\frac{\frac{7}{15}}{5x+3}\)
\(\Rightarrow (5x+3)(5x+3)=2\frac{1}{7}.\frac{7}{15}=1\)
\(\Leftrightarrow (5x+3)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} 5x+3=1\\ 5x+3=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{5}\\ x=-\frac{4}{5}\end{matrix}\right.\)