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|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|
\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)
\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)
\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)
Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}
|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0
|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|
\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)
Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
\(\frac{4}{7}x-\frac{2}{3}=\frac{1}{5}\)
\(\frac{4}{7}x=\frac{13}{15}\)
\(x=\frac{91}{60}\)
\(\frac{4}{5}+\frac{5}{7}:x=\frac{1}{6}\)
\(\frac{5}{7}:x=-\frac{19}{30}\)
\(x=-\frac{150}{133}\)
=.= hk tốt!!
câu 1b
Gọi d là ƯCLN (3n-7, 2n-5), d thuộc N*
Ta có : 3n-7 chia ht cho d , 2n_5 chia ht cho d
suy ra: 2(3n-7) chia ht cho d , 3(2n-5) chia ht cho d
suy ra 6n-14 chia ht cho d, 6n-15 chia ht cho d
dấu suy ra [(6n -15) - (6n-14)] chia ht cho d dấu suy ra 1 chia ht cho d suy ra d =1
Vậy......
1) b. Để chứng tỏ \(\frac{3n-7}{2n-5}\) là phân số tối giản
Ta cần chứng minh: ( 3n - 7; 2n - 5 ) = 1
Thật vậy: ( 3n - 7 ; 2n - 5 ) = ( 2n - 5 ; ( 3n - 7 ) - ( 2n - 5 ) ) = ( 2n - 5; n - 2 ) = ( n - 2; n - 3 ) = ( n - 2; 1 ) = 1
=> \(\frac{3n-7}{2n-5}\) là phân số tối giản
3) \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{12}\)
Ta có: \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\left(\frac{1}{5}+\frac{1}{7}\right)+\frac{1}{6}=\frac{12}{35}+\frac{1}{6}>\frac{12}{36}+\frac{1}{6}=\frac{2}{6}+\frac{1}{6}=\frac{1}{2}\)
\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}=\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}\right)>\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \)
=> A > 1/2 + 1/2 + 1/2 + 1/2 = 2
a) \(\frac{5}{6}=\frac{x-1}{x}\)
\(5x=6x-6\)
\(6x-5x=6\)
\(x=6\)
các câu còn lại lm tương tự
hok tốt!!
b) \(\frac{1}{2}=\frac{x+1}{3x}\)
\(\Rightarrow1.3x=2.\left(x+1\right)\)
\(3x=2x+2\)
\(3x-2x=2\)
\(x=2\)
Vậy x=2
các câu khác bạn làm tương tự
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
a) \(x+\frac{5}{12}=-1\frac{2}{7}\)
\(\Leftrightarrow x+\frac{5}{12}=\frac{-9}{7}\)
\(\Leftrightarrow x=\frac{-143}{84}\)
Vậy ...
b) \(4\frac{1}{2}x:\frac{5}{12}=0,5\)
\(\Leftrightarrow\frac{9}{2}x=\frac{5}{24}\)
\(\Leftrightarrow x=\frac{5}{108}\)
vậy...
c) \(7,5.1\frac{3}{4}x=6\frac{2}{5}\)
\(\Leftrightarrow\frac{105}{8}x=\frac{32}{5}\)
\(\Leftrightarrow x=\frac{256}{525}\)
Vậy ...
a) \(\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)\cdot\frac{11}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{7}\div\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=1\)
\(2x=\frac{9}{2}-1\)
\(x=\frac{7}{2}\div2\)
\(x=\frac{7}{4}\)
b) \(|\frac{3}{4}\cdot x-\frac{1}{2}|-1=\frac{1}{4}\)
\(|\frac{3}{4}\cdot x-\frac{1}{2}|=\frac{1}{4}+1\)
\(|\frac{3}{4}\cdot x|=\frac{5}{4}+\frac{1}{2}\)
\(x=\frac{7}{4}\div\frac{3}{4}\)
\(x=\frac{7}{3}\)
c) \(\frac{1}{4}-|3-x|=-\frac{3}{4}\)
\(|3-x|=\frac{1}{4}-\left(-\frac{3}{4}\right)\)
\(|3-x|=1\)
\(x=3-1\)
\(\Rightarrow x=2\)
d) \(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=1,4\)
\(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=\frac{7}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=\frac{7}{5}+\frac{3}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=2\)
\(\left(x-\frac{6}{7}\right)=2\div4\)
\(x=\frac{1}{2}+\frac{6}{7}\)
\(x=\frac{19}{14}\)
\(\)
Bài 1:
a; \(\dfrac{x}{3}\) = \(\dfrac{4}{y}\)
\(xy\) = 12
12 = 22.3; Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6;12}
Lập bảng ta có:
| \(x\) | -12 | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | 12 |
| y | -1 | -2 | -3 | -4 | -6 | -12 | 12 | 6 | 4 | 3 | 2 | 1 |
Theo bảng trên ta có các cặp \(x;y\) nguyên thỏa mãn đề bài là:
(\(x\)\(;y\)) =(-12; -1);(-6; -2);(-4; -3);(-2; -6);(-1; 12);(1; 12);(2;6);(3;4);(4;3);(6;2);(12;1)
b; \(\dfrac{x}{y}\) = \(\dfrac{2}{7}\)
\(x\) = \(\dfrac{2}{7}\).y
\(x\) \(\in\)z ⇔ y ⋮ 7
y = 7k;
\(x\) = 2k
Vậy \(\left\{{}\begin{matrix}x=2k\\y=7k;k\in z\end{matrix}\right.\)
\(\dfrac{2}{7}x-\dfrac{2}{5}=\dfrac{1}{4}\)
=>\(\dfrac{2}{7}x=\dfrac{2}{5}+\dfrac{1}{4}=\dfrac{8}{20}+\dfrac{5}{20}=\dfrac{13}{20}\)
=>\(x=\dfrac{13}{20}:\dfrac{2}{7}=\dfrac{13}{20}\cdot\dfrac{7}{2}=\dfrac{91}{40}\)
Chuyển \(-\frac25\) sang vế
\(\frac27\) x \(x\) \(=\frac14+\frac25\)
Tìm mẫu số chung của \(\frac14và\frac25\) để cộng hai phân số:
\(\frac14+\frac25\) \(=\frac{5}{20}+\frac{8}{20}=\frac{13}{20}\)
Phương trình trở thành:
\(\frac27\) x \(x\) \(=\) \(\frac{13}{20}\)
Nhân hai vế với nghịch đảo của \(\frac27\) thành \(\frac72\)
\(x=\) \(\frac{13}{20}\) x \(\frac72\)
\(x=\frac{91}{40}\)