X+5/x+7 + 3x+11/x+7
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a ) \(-5\times\left(-x+7\right)-3\times\left(-x-5\right)=-4\times\left(12-x\right)+48\)
\(\Leftrightarrow5x-35+3x+15=-48+4x+48\)
\(\Leftrightarrow5x-3x+4x=35-15-48+48\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
b ) \(-2\times\left(15-3x\right)-4\times\left(-7x+8\right)=-5-9\times\left(-2x+1\right)\)
\(\Leftrightarrow-30+6x-28x-32=-5+18x-9\)
\(\Leftrightarrow6x-28x-18x=30+32-5-9\)
\(\Leftrightarrow-40x=48\)
\(\Leftrightarrow x=-1.2\)
`x . 3/7 = 2/3`
`=>x= 2/3 : 3/7`
`=>x= 2/3 . 7/3`
`=>x=14/9`
`-----------`
`x : 8/11 = 11/3`
`=>x= 11/3 . 8/11`
`=>x= 88/33`
`=>x=8/3`
`-----------`
`4/7 . x - 2/3 = 1/5`
`=> 4/7 . x = 1/5 +2/3`
`=>4/7 . x =3/15 + 10/15`
`=>4/7 . x =13/15`
`=>x= 13/15 : 4/7`
`=>x= 13/15 xx 7/4`
`=>x= 91/60`
Lời giải:
$x.\frac{3}{7}=\frac{2}{3}$
$x=\frac{2}{3}: \frac{3}{7}=\frac{14}{9}$
-----------
$x: \frac{8}{11}=\frac{11}{3}$
$x=\frac{11}{3}.\frac{8}{11}=\frac{8}{3}$
-----------
$\frac{4}{7}x-\frac{2}{3}=\frac{1}{5}$
$\frac{4}{7}x=\frac{2}{3}+\frac{1}{5}=\frac{13}{15}$
$x=\frac{13}{15}: \frac{4}{7}=\frac{91}{60}$
\(a,-5.\left(-x+7\right)-3.\left(-x-5\right)=-4.\left(12-x\right)+48\)
\(5x-35-3x+15=-48+4x+48\)
\(2x-10=4x\)
\(2x-4x=10\)
\(-2x=10\)
\(x=-5\)
\(b,\left(-x-7\right)-5.\left(-x-3\right)=12.\left(3-x\right)\)
\(-x-7+5x+15=36-12x\)
\(4x+8=36-12x\)
\(4x+12x=36-8\)
\(16x=28\)
\(x=1,75\)
các câu còn lại tương tự nha
b)x+3=4:2
=> x=-1
d)5x-15=3x-5
<=> 5x-3x=15-5
<=> 2x=10
<=> x=5
f) 35-7x=11-5x
<=> 35-11=-5x+7x
<=> 24=2x
<=> x=12
h) 6x-2-3x=10
<=> 3x=10+2
<=> x=4
j)3-2x=3x+3-x-2
<=> 3-2x=2x+1
<=>-4x=-2
<=> x=1/2
Hướng dẫn cách làm:
B1:Bạn phá ngoặc ra
B2:Bạn chuyển hết x sang một bên(Nhớ sử dụng quy tắc chuyển vế đổi dấu)
B3:Bạn tính hai vế
(Áp dụng quy tắc chuyển vế đổi dấu)
(Hoặc)
B1:Bạn tính hai vế
B2:Chuyển hết x sang một vế
B3:Tính vế chứa x
(Áp dụng quy tắc lấy tiềnmở khóa cổng-mở khóa cửa-mở khóa két-thò tay vào lấy tiền)
P/s:Khi tìm x nhớ áp dụng hai quy tắc mình nói trên
a) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(2x-3=4\)
\(2x=7\)
\(x=\dfrac{7}{2}=3,5\)
b) \(\left(3x-2\right)^5=-243\)
\(\left(3x-2\right)^5=-3^5\)
\(3x-2=-3\)
\(3x=-1\)
\(3x=-\dfrac{1}{3}\)
c) \(\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\left(x-7\right)^{x+1}\times\left[1-\left(x-7\right)^{10}\right]=0\)
\(\left(x-7\right)^{x+1}=0\) ; \(1-\left(x-7\right)^{10}=0\)
\(x-7=0;\left(x-7\right)^{10}=1\)
\(x=7;\left(x-7=1;x-7=-1\right)\)
\(x=7;x=8;x=6\)
a, (2\(x\) - 3)2 = 16
\(\left[{}\begin{matrix}2x-3=-4\\2x-3=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-1\\2x=7\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(x\in\){ - \(\dfrac{1}{2}\); \(\dfrac{7}{2}\)}
b, (3\(x\) - 2)5 = -243
( 3\(x\) - 2)5 = (-3)5
3\(x\) - 2 = -3
3 \(x\) = -1
\(x\) = - \(\dfrac{1}{3}\)
Vậy \(x\) = -\(\dfrac{1}{3}\)
c, \(\left(x-7\right)\)\(x+1\) = (\(x-7\))\(x+11\)
(\(x-7\))\(^{x+1}\).( \(\left(x-7\right)^{10}\) - 1 ) = 0
\(\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x-7=-1\\x-7=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)
Vậy \(x\in\){ 6; 7; 8}
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)
=>7/6x=2/3
hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)
hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
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