Tìm x: \(\frac{4X+7}{3}=\frac{12}{4x+7}\)
Những câu hỏi liên quan
Tìm x:\(a,\frac{6x-5}{-7}=\frac{5x-3}{-5}\\ b,\frac{12-7x}{-13}=\frac{4-3x}{-5}\\ c,\frac{2x+4}{7}=\frac{4x-2}{15}\)
a) \(\frac{6x-5}{-7}=\frac{5x-3}{-5}\)
=> -5(6x - 5) = -7(5x - 3)
=> -30x + 25 = -35x + 21
=> -30x + 25 + 35x - 21 = 0
=> (-30x + 35x) + (25 - 21) = 0
=> 5x + 4 = 0
=> 5x = -4
=> x = -4/5
b) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)
=> -5(12 - 7x) = -13(4 - 3x)
=> -60 + 35x = -52 + 39x
=> -60 + 35x + 52 - 39x = 0
=> (-60 + 52) + (35x - 39x) = 0
=> -8 - 4x = 0
=> -8 = 4x
=> x = -2
c) \(\frac{2x+4}{7}=\frac{4x-2}{15}\)
=> 15(2x + 4) = 7(4x - 2)
=> 30x + 60 = 28x - 14
=> 30x + 60 - 28x + 14 = 0
=> 2x + 74 = 0
=> 2x = -74
=> x = -37
Tìm GTNN
A=\(\frac{21|4x+6|+33}{3|4x+6|+5}\)
B=\(\frac{6|y+5|+14}{2|+5|+14}\)
C=\(\frac{-15|x+7|-68}{3|x+7|+12}\)
$\frac{4x+3}{5}$ -$\frac{6x-2}{7}$ =$\frac{5x+4}{3}$ +3
b.
$\frac{x+4}{5}$ -x+4=$\frac{x}{3}$ -$\frac{x-2}{2}$
c.$\frac{5x+2}{6}$ -$\frac{8x-1}{3}$ =$\frac{4x+2}{5}$ -5
d.$\frac{2x+3}{3}$ =$\frac{5-4}{2}$
e. $\frac{5x+3}{12}$ =$\frac{1+2x}{9}$
f.$\frac{7x-1}{6}$ =$\frac{16-x}{5}$
g. $\frac{x-3}{5}$ =6-$\frac{1-2x}{3}$
h. $\frac{3x-2}{6}$ -5=$\frac{3-2(x+7)}{4}$
giúp vs ạ, cần gấp
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
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\(\frac{x-1}{12}+\frac{2x-12}{14}=\frac{3x-14}{25}+\frac{4x-29}{27}\)\(\frac{3}{x+1}+\frac{4}{x-2}=/frac{5}{x-3}+\frac{6}{x-4}\)\(\frac{2x-1}{2}+\frac{2x+1}{3}=/frac{2x+7}{6}+\frac{2x+9}{7}\)
Xem chi tiết
tìm x biết \(\frac{4x}{4x^2-8x+7}+\frac{3x}{4x^2-10x+7}=1\)
tìm A, B, C, D
a, \(\frac{64x^3+1}{16x^2-2}=\frac{A}{4x-1}\)
b, \(\frac{4x^2+3x-7}{B}=\frac{4x+7}{2x-3}\)
c, \(\frac{C}{3x^2-7x+4}=\frac{3-2x}{x-\frac{4}{3}}\)
d, \(\frac{2x-y-1}{4x-2y}=\frac{4x^2-2x-y^2-y}{D}\)
\(\frac{x-1}{12}+\frac{2x-3}{23}+\frac{3x-5}{34}+\frac{4x-7}{45}-4\)
\(=\left(\frac{x-1}{12}-1\right)+\left(\frac{2x-3}{23}-1\right)+\left(\frac{3x-5}{34}-1\right)+\left(\frac{4x-7}{45}-1\right)\)
\(=\frac{x-13}{12}+2.\frac{x-13}{23}+3.\frac{x-13}{34}+4.\frac{x-13}{45}\)
\(=\left(x-13\right)\left(\frac{1}{12}+\frac{2}{23}+\frac{3}{34}+\frac{4}{45}\right)\)
Đề thiếu phải không??? :))
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Nguyễn Lam Giang
Tìm x mà không có vế phải thì tìm bằng niềm tin ak
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Xem thêm câu trả lời
\(\frac{1}{3.7}+\frac{1}{7.11}+.....+\frac{1}{\left(4x+3\right)\left(4x+7\right)}=\frac{5}{12}\)
\(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{\left(4x+3\right)\left(4x+7\right)}=\frac{5}{12}\)(x phải khác \(-\frac{3}{4};-\frac{7}{4}\)nhé)
\(\Leftrightarrow\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4x+3\right)\left(4x+7\right)}=4.\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4x+3}-\frac{1}{4x+7}=\frac{5}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{4x+7}=\frac{5}{3}\)
\(\Leftrightarrow\frac{4x+7-3}{3\left(4x+7\right)}=\frac{5\left(4x+7\right)}{3\left(4x+7\right)}\)
\(\Rightarrow4x+7-3=20x+35\)(chỗ này dùng dấu suy ra nhé)
\(\Leftrightarrow4x-20x=35-7+3\)
\(\Leftrightarrow-16x=31\)
\(\Leftrightarrow x=-\frac{31}{16}\)
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Cho biết \(B=\frac{4x+4}{2004}+\frac{4x+5}{2003}=\frac{4x+6}{2002}+\frac{4x+7}{2001}\)Tìm x