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Đặt \(A=1+2+2^2+2^3+\cdots+2^{2023}\)
=>\(2A=2+2^2+2^3+2^4+\cdots+2^{2024}\)
=>\(2A-A=2+2^2+2^3+2^4+\cdots+2^{2024}-1-2-\cdots-2^{2023}\)
=>\(A=2^{2024}-1\)
Ta có: \(2^{2024}-\left(1+2+2^2+2^3+\cdots+2^{2023}\right)\)
\(=2^{2024}-\left(2^{2024}-1\right)=1\)
2^ 2024 -( 1 + 2 + 2 ^ 2 + 2 ^ 3 +...+2^ 2023 )
2^2024 -(2^2024-1)
1
Nhưng dài lắm chắc lớp 6 ko hiểu đc nên tui chỉ viết ngắn gọn
BẠN THÔNG CẢM
/-x/<5
=> x thuộc { 0;1;-1;2;-2;3;-3;4;-4 }
Tổng các số nguyên trên là :
= [(-1)+1]+[(-2)+2]+[(-3)+3]+[(-4)+4]+0
=0+0+0+0+0
=0.
#Hoc tot.
Câu a: 2\(^3\).\(x\) : |\(\frac{13}{23}\) - 5\(\frac{13}{23}\)| = - 2\(\frac13\)
\(8x\) : |-5| = - \(\frac73\)
8\(x\) : 5 = -\(\frac73\)
8\(x\) = - \(\frac73\times5\)
8\(x\) = - \(\frac{35}{3}\)
\(x\) = - \(\frac{35}{3}\) : 8
\(x\) = - \(\frac{35}{24}\)
Vậy \(x=-\frac{35}{24}\)
Câu b:
||\(x-3\)| - 5| = 7
|\(x-3\)| - 5 = 7 hoặc |\(x-3\)| - 5 = - 7
TH1: |\(x-3\)| - 5 = 7
|\(x-3\)| = 7+ 5
|\(x-3\)| = 12
\(\left[\begin{array}{l}x-3=-12\\ x-3=12\end{array}\right.\)
\(\left[\begin{array}{l}x=-12+3\\ x=12+3\end{array}\right.\)
\(\left[\begin{array}{l}x=-9\\ x=15\end{array}\right.\)
TH2: |\(x-3\)| - 5 = - 7
|\(x-3\)| = - 7 + 5
|\(x\) - 3| = - 2(loại) Vì trị tuyêt đối của một số luôn không âm.
Vậy \(x\in\) {-9; 15}
A = \(\frac{2^3.3}{2^2.3^2.5}\)
A = \(\frac{2^2.3.2}{2^2.3.3.5}\)
A = \(\frac{2}{3.5}\)
A = \(\frac{2}{15}\)
\(\frac{2^3.3}{2^2.3^2.5}=\frac{2}{3.5}=\frac{2}{15}\)
Thiếu dấu nhân ở chỗ \(2^2.3^2\)nha
(7x - 11)3 = 25 . 52 + 200
(7x - 11)3 = 32 . 25 + 200
(7x - 11)3 = 800 + 200
(7x - 11)3 = 1000
(7x - 11)3 = 103
=> 7x - 11 = 10
=> 7x = 10 + 11
=> 7x = 21
=> ko có giá trị nào của x thỏa mãn
Ủng hộ mk nha !!! ^_^
a: Ta có: \(A=2+2^2+2^3+\cdots+2^{2025}\)
=>\(2A=2^2+2^3+2^4+\cdots+2^{2026}\)
=>\(2A-A=2^2+2^3+2^4+\cdots+2^{2026}-2-2^2-\cdots-2^{2025}\)
=>\(A=2^{2026}-2\)
b:Sửa đề: \(B=1+5+5^2+\cdots+5^{150}\)
=>\(5B=5+5^2+5^3+\cdots+5^{151}\)
=>\(5B-B=5+5^2+5^3+\cdots+5^{151}-1-5-5^2-\cdots-5^{150}\)
=>\(4B=5^{151}-1\)
=>\(B=\frac{5^{151}-1}{4}\)
c: Ta có: \(C=3+3^2+3^3+\ldots+3^{1000}\)
=>\(3C=3^2+3^3+3^4+\cdots+3^{1001}\)
=>\(3C-C=3^2+3^3+\cdots+3^{1001}-3-3^2-\cdots-3^{1000}\)
=>\(2C=3^{1001}-3\)
=>\(C=\frac{3^{1001}-3}{2}\)
a,\(\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x};Đkxđ:x\ne1\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}\left(\frac{-9}{20}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{2-2x}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{-2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{7}{2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow70=-6\left(x-1\right)\)
\(\Rightarrow6x=6-70\)
\(\Rightarrow6x=-64\)
\(\Rightarrow x=\frac{-32}{3}x\ne1\)
Đề bài:
\(A = \frac{1 2^{3} \cdot 12 1^{2} \cdot 5 - 2 2^{4} \cdot 3^{3}}{7 5^{2} \cdot 1 1^{4} - 3 0^{2} \cdot 1 1^{5}}\)
Bước 1: Biến đổi các lũy thừa
Ta rút gọn từng số:
- \(1 2^{3} = \left(\right. 3 \cdot 4 \left.\right)^{3} = 3^{3} \cdot 4^{3} = 3^{3} \cdot \left(\right. 2^{2} \left.\right)^{3} = 3^{3} \cdot 2^{6}\)
- \(12 1^{2} = \left(\right. 1 1^{2} \left.\right)^{2} = 1 1^{4}\)
- \(2 2^{4} = \left(\right. 2 \cdot 11 \left.\right)^{4} = 2^{4} \cdot 1 1^{4}\)
- \(3^{3} = 3^{3}\)
- \(7 5^{2} = \left(\right. 3 \cdot 5^{2} \left.\right)^{2} = 3^{2} \cdot 5^{4}\)
- \(1 1^{4} = 1 1^{4}\)
- \(3 0^{2} = \left(\right. 2 \cdot 3 \cdot 5 \left.\right)^{2} = 2^{2} \cdot 3^{2} \cdot 5^{2}\)
- \(1 1^{5} = 1 1^{5}\)
Bước 2: Thay vào biểu thức
Tử số:
\(1 2^{3} \cdot 12 1^{2} \cdot 5 - 2 2^{4} \cdot 3^{3} = \left(\right. 3^{3} \cdot 2^{6} \cdot 1 1^{4} \cdot 5 \left.\right) - \left(\right. 2^{4} \cdot 1 1^{4} \cdot 3^{3} \left.\right)\)
Rút chung:
\(= 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 2^{2} \cdot 5 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 4 \cdot 5 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot \left(\right. 20 - 1 \left.\right) = 3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot 19\)
Mẫu số:
\(7 5^{2} \cdot 1 1^{4} - 3 0^{2} \cdot 1 1^{5} = \left(\right. 3^{2} \cdot 5^{4} \cdot 1 1^{4} \left.\right) - \left(\right. 2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 1 1^{5} \left.\right)\)
Rút chung \(3^{2} \cdot 5^{2} \cdot 1 1^{4}\):
\(= 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 5^{2} - 2^{2} \cdot 11 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 25 - 4 \cdot 11 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. 25 - 44 \left.\right) = 3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. - 19 \left.\right)\)
Bước 3: Viết lại biểu thức đầy đủ
\(A = \frac{3^{3} \cdot 2^{4} \cdot 1 1^{4} \cdot 19}{3^{2} \cdot 5^{2} \cdot 1 1^{4} \cdot \left(\right. - 19 \left.\right)}\)
Rút gọn:
- \(3^{3} / 3^{2} = 3\)
- \(1 1^{4}\) triệt tiêu
- \(19 / \left(\right. - 19 \left.\right) = - 1\)
Còn lại:
\(A = \frac{3 \cdot 2^{4}}{5^{2}} \cdot \left(\right. - 1 \left.\right) = \frac{3 \cdot 16}{25} \cdot \left(\right. - 1 \left.\right) = \frac{48}{25} \cdot \left(\right. - 1 \left.\right) = \boxed{- \frac{48}{25}}\)
✅ Kết quả cuối cùng:
\(\boxed{A = - \frac{48}{25}}\)
Ta có: \(12^3\cdot121^2\cdot5-22^4\cdot3^3\)
\(=\left(2^2\cdot3\right)^3\cdot\left(11^2\right)^2\cdot5-11^4\cdot2^4\cdot3^3\)
\(=2^6\cdot3^3\cdot11^4\cdot5-11^4\cdot2^4\cdot3^3=11^4\cdot3^3\cdot2^4\left(2^2\cdot5-1\right)\)
\(=11^4\cdot3^3\cdot2^4\cdot19\)
Ta có: \(75^2\cdot11^4-30^2\cdot11^5\)
\(=\left(3\cdot5^2\right)^2\cdot11^4-\left(2\cdot3\cdot5\right)^2\cdot11^5\)
\(=3^2\cdot5^4\cdot11^4-2^2\cdot3^2\cdot5^2\cdot11^5\)
\(=3^2\cdot5^2\cdot11^4\left(5^2-2^2\cdot11\right)=3^2\cdot5^2\cdot11^4\cdot\left(-19\right)\)
Ta có: \(A=\frac{12^3\cdot121^2\cdot5-22^4\cdot3^3}{75^2\cdot11^4-30^2\cdot11^5}\)
\(=\frac{2^4\cdot3^3\cdot11^4\cdot19}{3^2\cdot5^2\cdot11^4\cdot\left(-19\right)}=\frac{2^4\cdot3}{5^2\cdot\left(-1\right)}=\frac{48}{-25}\)
Ta có: \(A=1+2+2^2+2^3+\cdots+2^{2024}+2^{2025}\)
\(\) \(\) \(2A=2+2^2+2^3+2^4+\cdots+2^{2025}+2^{2026}\)
\(2A-A=\left(2+2^2+2^3+\cdots+2^{2025}+2^{2026}\right)-\left(1+2+2^2+2^3+\cdots+2^{2024}+2^{2025}\right)\)
\(2A=2^{2026}-1\)
ko phải cho dc 5 sao đâu
\(2A=2+2^2+2^3+2^4+\ldots+2^{2025}+2^{2026}\)
\(2A-A=\left(\right.2+2^2+2^3+\ldots+2^{2025}+2^{2026}\left.\right)-\left(\right.1+2+2^2+2^3+\ldots+2^{2024}+2^{2025}\left.\right)\)
\(2 A = 2^{2026} - 1\)
Ta có: \(A = 1 + 2 + 2^{2} + 2^{3} + \hdots + 2^{2024} + 2^{2025}\)
\(\) \(\) \(2 A = 2 + 2^{2} + 2^{3} + 2^{4} + \hdots + 2^{2025} + 2^{2026}\)
\(2 A - A = \left(\right. 2 + 2^{2} + 2^{3} + \hdots + 2^{2025} + 2^{2026} \left.\right) - \left(\right. 1 + 2 + 2^{2} + 2^{3} + \hdots + 2^{2024} + 2^{2025} \left.\right)\)
\(2 A = 2^{2026} - 1\)
Ta có: \(A=1+2+2^2+2^3+\cdots+2^{2024}+2^{2025}\)
\(\) \(\) \(2A=2+2^2+2^3+2^4+\ldots+2^{2025}+2^{2026}\)
\(2A-A=\left(\right.2+2^2+2^3+\cdots+2^{2025}+2^{2026}\left.\right)-\left(\right.1+2+2^2+2^3+\ldots+2^{2024}+2^{2025}\left.\right)\)
\(2 A = 2^{2026} - 1\)