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Bài 1:
a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)
\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)
\(=-\frac{3}{7}\)
b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)
\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)
\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)
\(=\frac{1}{2}-\frac{4}{5}\)
\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)
d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)
\(=5^2+2^5-\frac{15^2}{15^2}\)
\(=25+32-1\)
\(=56\)
e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)
\(=\frac{4}{17}+\frac{13}{17}\)
\(=\frac{17}{17}=1\)
g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)
\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)
\(=\frac{7}{12}\cdot4=\frac{7}{3}\)
Ta có: B = 22010 - 22009 - 22008 -......- 2 -1
=> B = 22010 - (1 + 2 + 22 + ..... + 22009)
Đặt A = 1 + 2 + 22 + .... + 22009
=> 2A = 2 + 22 + .... + 22010
=> 2A - A = 22010 - 1
=> A = 22010 - 1
Vậy B = 22010 - (22010 - 1)
=> B = 22010 - 22010 + 1
=> B = 1
Ta có: B = 22010 - 22009 - 22008 -......- 2 -1
=> B = 22010 - (1 + 2 + 22 + ..... + 22009)
Đặt A = 1 + 2 + 22 + .... + 22009
=> 2A = 2 + 22 + .... + 22010
=> 2A - A = 22010 - 1
=> A = 22010 - 1
Vậy B = 22010 - (22010 - 1)
=> B = 22010 - 22010 + 1
=> B = 1
\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}x\) = \(\frac{1}{10}+\frac{1}{2}\)
\(\frac{2}{3}x\) = \(\frac{3}{5}\)
\(x\) = \(\frac{3}{5}:\frac{2}{3}\)
\(x\) = \(\frac{9}{10}\)
Câu b:
(2 4/5x - 50) : 2/3 = 51
2 4/5x - 50 = 51 x 2/3
14/5x - 50 = 34
14/5x = 34 + 50
14/5x = 84
x = 84 : 14/5
x = 30
Vậy x = 30
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
1. a) 2B = 1 + 1/2 + 1/22+...+1/298
B - B = (1+1/2+...+1/298) - (1/2+....+1/299)
B = 1 - 299 => B < 1
b) Làm tương tự như câu a, ra là (1 - 1/399) : 2 = 1/2 - 1/2.399(C bé hơh 1/2)
1. a). Theo đầu bài ta có:
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(\Leftrightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(\Leftrightarrow B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{2^{99}}< 1\)( đpcm )

(x-1)2=|-1|
=> (x-1)2=1
=>x-1=1
=>x=2
thanks bạn nha!
Bạn Top 10 Gunny thiếu 1 trường hợp rồi nhé
\(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=\left|-1\right|\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=1\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
Vậy \(x=0\) hoặc \(x=2\)
Chúc bạn học tốt ~
(x-1)2=|-1|
=>(x-1)^2=1
=>x-1=1
x=1+1
x=2
bạn Phùng Minh Quân sai rùi nhé bạn thử lại xem
\(x-1 2 = 4 1 − 2 1 − 4 3 ⇔ x − 1 2 = −1 ⇔ x − 1 2 = 1 ⇔ x − 1 = 1 x − 1 = −1 ⇔ x = 2 x = 0 Vậy x = 0 hoặc x \)
x − 1 2 = 4 1 − 2 1 − 4 3 ⇔ x − 1 2 = −1 ⇔ x − 1 2 = 1 ⇔ x − 1 = 1 x − 1 = −1 ⇔ x = 2 x = 0
Vậy x = 0 hoặc x