\(\frac{2x-3}{x+1}=\frac{21}{16}\)
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NT
2
10 tháng 3 2020
a, \(\frac{3}{x-1}\) = \(\frac{21}{16}\)
=> 3 . 16 = 21 . ( x - 1 ) = 48
x - 1 = 48 : 21 = \(\frac{16}{7}\)
x = \(\frac{16}{7}\) + 1 = \(\frac{23}{7}\)
b, \(\frac{2x-3}{x-2}\) = \(\frac{3}{4}\)
=> 4 . ( 2x - 3 ) = 3 . ( x - 2 )
=> 8x - 12 = 3x - 6
=> 8x -12 - 3x + 6 = 0
= 5x - 6 = 0; 5x =6
=> x = \(\frac{6}{5}\)
30 tháng 4 2019
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};3\right\}\)
e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
30 tháng 4 2019
Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
TT
12 tháng 8 2020
\(-\frac{17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)
\(\Leftrightarrow-\frac{17}{21}:\frac{17}{20}< x+\frac{4}{7}< \frac{12}{12}-\frac{6}{12}+\frac{4}{12}-\frac{3}{12}\)
\(\Leftrightarrow-\frac{17}{21}.\frac{20}{17}< x+\frac{4}{7}< \frac{7}{12}\)
\(\Leftrightarrow-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)
\(\Leftrightarrow-\frac{20}{21}< x< \frac{1}{84}\)
\(\Leftrightarrow-\frac{80}{84}< x< \frac{1}{84}\)
\(\Leftrightarrow-80< x< 1\Leftrightarrow x\in\left\{-79;-78;...;0\right\}\)
mà để Giá trị nguyên lớn nhất của x
\(\Rightarrow x=-1\)
6 tháng 9 2020
f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)( ĐKXĐ : \(x\ne-\frac{1}{2}\))
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x^2=\left(\pm4\right)^2\)
\(\Leftrightarrow x=\pm4\)(tmđk)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)( ĐKXĐ : \(x\ne-1\))
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=6\cdot5\)
\(\Leftrightarrow10x^2+15x+5=30\)
\(\Leftrightarrow10x^2+15x+5-30=0\)
\(\Leftrightarrow10x^2+15x-25=0\)
\(\Leftrightarrow5\left(2x^2+3x-5\right)=0\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)(tmđk)
NN
6 tháng 9 2020
f) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=21.3\)
\(\Leftrightarrow4x^2-1=63\)
\(\Leftrightarrow4x^2=64\)
\(\Leftrightarrow x^2=16\)\(\Leftrightarrow x^2=4^2\)\(\Leftrightarrow x=4\)
Vậy \(x=4\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\)
\(\Leftrightarrow\left(10x+5\right)\left(x+1\right)=5.6\)
\(\Leftrightarrow5\left(2x+1\right)\left(x+1\right)=30\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=6\)
\(\Leftrightarrow2x^2+3x+1=6\)
\(\Leftrightarrow2x^2+3x-5=0\)
\(\Leftrightarrow\left(2x^2-2x\right)+\left(5x-5\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{-5}{2};1\right\}\)
TG
30 tháng 3 2020
a/ 2x = 5y và x - 2y = -12
Ta có: 2x = 5y => \(\frac{x}{5}=\frac{y}{2}\)
Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{5}=\frac{y}{2}=\frac{x-y}{5+2}=\frac{x-2y}{5+2.2}=\frac{-12}{9}=-\frac{4}{3}\)
\(\frac{x}{5}=-\frac{4}{3}\Rightarrow x=\frac{-4}{3}.5=-\frac{20}{3}\)
\(\frac{y}{2}=-\frac{4}{3}\Rightarrow y=-\frac{4}{3}.2=-\frac{8}{3}\)
Vậy:.................
b/ 2x = 3y = 4z và x + y + z =21
Ta có: 2x = 3y = 4z
=> \(\frac{2x}{12}=\frac{3y}{12}=\frac{4z}{12}\)
=> \(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)
Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{6+4+3}=\frac{21}{13}\)
\(\frac{x}{6}=\frac{21}{13}\Rightarrow x=\frac{21}{13}.6=\frac{126}{13}\)
\(\frac{y}{4}=\frac{21}{13}\Rightarrow y=\frac{21}{13}.4=\frac{84}{13}\)
\(\frac{z}{3}=\frac{21}{13}\Rightarrow z=\frac{21}{13}.3=\frac{63}{13}\)
Vậy:...............
c/Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{32}{8}=4\)
\(\frac{x}{3}=4\Rightarrow x=4.3=12\)
\(\frac{y}{5}=4\Rightarrow y=4.5=20\)
Vậy:................
d/ Ta có: 7x = 3y
=> \(\frac{7x}{21}=\frac{3y}{21}\)
=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)
\(\frac{x}{4}=-4\Rightarrow x=\left(-4\right).4=-16\)
\(\frac{y}{7}=-4\Rightarrow y=\left(-4\right).7=-28\)
Vậy:................
KN
18 tháng 8 2020
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
19 tháng 8 2020
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
13 tháng 2 2019
Làm đc 2 bài đầu chưa, t làm câu cuối cho, hai câu đầu dễ í mà
\(\frac{2x-3}{x+1}=\frac{21}{16}\)
\(\left(2x-3\right)\cdot16=21\left(x+1\right)\)
\(32x-48=21x+21\)
\(32x-21x=21+48\)
\(11x=69\)
\(x=\frac{69}{11}\)