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a) Ta có: \(71+\frac{26-3x}{5}=75\)
\(\Leftrightarrow\frac{26-3x}{5}=4\)
\(\Leftrightarrow26-3x=20\)
\(\Leftrightarrow3x=6\)
hay x=2
Vậy: x=2
b) Ta có: \(\left|x-12\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-12=5\\x-12=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-5+12=7\end{matrix}\right.\)
Vậy: \(x\in\left\{17;7\right\}\)
c) Ta có: \(\frac{7}{8}+x=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}-\frac{7}{8}\)
\(\Leftrightarrow x=\frac{24}{40}-\frac{35}{40}\)
hay \(x=-\frac{11}{40}\)
Vậy: \(x=-\frac{11}{40}\)
d) Ta có: \(\frac{1}{2}x-\frac{2}{5}=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2\)
hay \(x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
B1
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)
\(x-\frac{11}{6}=1:\frac{3}{50}\)
\(x-\frac{11}{6}=\frac{50}{3}\)
\(x=\frac{50}{3}+\frac{11}{6}\)
\(x=\frac{37}{2}\)
b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)
\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)
\(\frac{5}{7}:x=-\frac{4}{15}\)
\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)
\(x=-\frac{75}{28}\)
c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)
\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)
\(\frac{2}{5}.x=-1\)
\(x=-1:\frac{2}{5}\)
\(x=-\frac{5}{2}\)
B2
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)
\(=\frac{7}{6}.24:5-\frac{33}{10}\)
\(=28:5-\frac{33}{10}\)
\(=\frac{28}{5}-\frac{33}{10}\)
\(=\frac{56}{10}-\frac{33}{10}\)
\(=\frac{23}{10}\)
b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)
\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)
\(=\frac{61}{70}+0:\frac{25}{144}\)
\(=\frac{61}{70}+0\)
\(=\frac{61}{70}\)
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+....+\frac{2}{x}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{6}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{3}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{3}{9}-\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
câu a khó quá.Để nghĩ.
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{21\cdot2}+\frac{2}{28\cdot2}+\frac{2}{36\cdot2}+.....+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\left(x-1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)
\(\Rightarrow9\left(x-5\right)=6x+6\)
\(\Rightarrow9x-45=6x+6\)
\(\Rightarrow9x-6x=51\)
\(\Rightarrow3x=51\)
Tới đây bí:v
a) \(22\frac{1}{2}\cdot\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}\cdot\frac{7}{9}+\frac{50}{100}-\frac{125}{100}\)
\(=\frac{5}{2}\cdot\frac{7}{1}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}=18-\frac{5}{4}=\frac{67}{4}\)
b) \(1,4\cdot\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(=\frac{7}{5}\cdot\frac{15}{49}-\frac{22}{15}:\frac{11}{15}\)
\(=\frac{1}{1}\cdot\frac{3}{7}-\frac{22}{15}\cdot\frac{15}{11}\)
\(=\frac{3}{7}-2=\frac{3-14}{7}=\frac{-11}{7}\)
c) \(\left(-\frac{1}{2}\right)^2-\frac{7}{16}:\frac{7}{4}+75\%\)
\(=\frac{1}{4}-\frac{7}{16}\cdot\frac{4}{7}+\frac{75}{100}\)
\(=\frac{1}{4}-\frac{1}{4}+\frac{3}{4}=\frac{3}{4}\)
Bài 2 Bạn tự làm nhé
1.a,\(22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{67}{4}\)
b,Các phép tính khác làm tương tự
Đổi các số ra hết thành phân số,có ngoặc thì lm ngoặc trc,Xoq đến nhân chia trước dồi mới cộng trừ
c,tương tự
2.
a,\(1\frac{3}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{8}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{7}{12}\div x=\frac{-77}{20}\)
Đến đây dễ bạn tự làm
b,\(\left(2\frac{4}{5}.x+50\right)\div\frac{2}{3}=-51\)
\(\left(\frac{14}{5}x+50\right)\div\frac{2}{3}=-51\)
\(\frac{14}{5}x+50=-34\)
\(\frac{14}{5}x=-84\)
Tự làm tiếp
c,\(\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)\(\Rightarrow\left|\frac{3}{4}x-\frac{1}{2}\right|=\varnothing\)
Ta có: \(\frac{12}{36}=\frac{1}{3}\)mà \(\frac{12}{36}=\frac{2}{3\times x}\)
\(\Rightarrow\frac{1}{3}=\frac{2}{3\times x}\)
Vậy x = (2 : 1) x 3 = 6
Vậy x = 6
Mấy câu kia TT