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Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
\(a,\dfrac{5}{8}=\dfrac{x}{14}\)
\(\Rightarrow x=\dfrac{5.14}{8}=8,75\)
Vậy \(x=8,75\)
\(b,\dfrac{x}{6}=-\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{1.6}{3}=-2\)
Vậy \(x=-2\)
\(c,-\dfrac{3}{5}=\dfrac{x}{10}\)
\(\Rightarrow x=-\dfrac{3.10}{5}=-6\)
Vậy \(x=-6\)
câu d đã có đáp án
a: =>(2x+1/2)^2=1/4
=>2x+1/2=1/2 hoặc 2x+1/2=-1/2
=>x=-1/2 hoặc x=0
b: =>(x-1/5)^2=49
=>x-1/5=7 hoặc x-1/5=-7
=>x=-6,8 hoặc x=7,2
c: =>1,2x=12
=>x=10
d: =>3/4x+1/2x+1/2=-11/4
=>5/4x=-11/4-2/4=-13/4
=>x=-13/5
e: =>-0,25x+1,25x=0,2
=>x=0,2
a: \(\Leftrightarrow\left(\dfrac{13}{4}:x\right)\cdot\left(-\dfrac{5}{4}\right)=\dfrac{-10}{6}-\dfrac{5}{6}=\dfrac{-15}{6}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{13}{4}:x=\dfrac{5}{2}\cdot\dfrac{5}{4}=\dfrac{25}{8}\)
hay \(x=\dfrac{13}{4}:\dfrac{25}{8}=\dfrac{13}{4}\cdot\dfrac{8}{25}=\dfrac{26}{25}\)
b: \(\Leftrightarrow\dfrac{3}{4}:x=\dfrac{11}{36}-\dfrac{1}{4}=\dfrac{2}{36}=\dfrac{1}{18}\)
=>\(x=\dfrac{3}{4}:\dfrac{1}{18}=\dfrac{54}{4}=\dfrac{27}{2}\)
c: \(\Leftrightarrow\left(-\dfrac{6}{5}+x\right):\left(-3.6\right)=-\dfrac{7}{4}+\dfrac{1}{4}\cdot8=\dfrac{1}{4}\)
=>x-6/5=-9/10
=>x=3/10
\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{15}\)
3 \(\times\) y = 2 - \(\dfrac{8}{15}\)
3 \(\times\) y = \(\dfrac{22}{15}\)
y = \(\dfrac{22}{15}\) : 3
y = \(\dfrac{22}{45}\)
Câu 6: Khôg có cau nào đúng
Câu 7: C
Câu 8: B
Câu 9: B
Câu 10: D
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)