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1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
| 2x+1 | 1 | 3 | 7 | 21 |
| x | 0 | 1 | 3 | 10 |
| TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.
\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)
\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)
\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)
\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)
\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\Rightarrow-\frac{1}{12}\left(x-2018\right)=0\Leftrightarrow x=2018\)
Bài làm :
Ta có :
\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)
\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)
\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)
\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)
\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\text{Vì : }\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\ne0\Rightarrow x-2018=0\)
\(\Rightarrow x=2018\)
Vậy x=2018
Tìm x:
b) 1/3.x+2/5.(x-1)=0
\(<=> \dfrac{1}{3} .x +\dfrac{2}{5}x - \dfrac{2}{5} =0\)
\(<=> \dfrac{11}{15}x = \dfrac{2}{5}\)
\(<=> x= \dfrac{6}{11}\)
Vậy \( x= \dfrac{6}{11}\)
c) (2x-3).(6-2x)=0
\(<=> \begin{cases}
2x-3=0 \\
6-2x=0
\end{cases}\) \(<=> \begin{cases}
2x=3 \\
-2x=-6
\end{cases}\) \(<=>\begin{cases}
x=\dfrac{3}{2} \\
x=3
\end{cases}\)
Vậy \(x=( \dfrac{3}{2} ; 3)\)
d) -2/3-1/3.(2x-5)= 3/2
\(<=> 2x-5= \dfrac{5}{2}\)
\(<=> 2x= \dfrac{15}{2}\)
\(<=> x= \dfrac{15}{4}\)
Vậy \(x= \dfrac{15}{4}\)
f) 1/3.x-1/2=4 và 1/2 (Hỗn số ý '^')
\(<=> \dfrac{1}{3} x -\dfrac{1}{2} = \dfrac{9}{2}\)
\(<=> \dfrac{1}{3}x =5\)
\(<=> x= 15\)
Vậy \(x= 15\)
\(\left(x+1\right)\left(x+7\right)< 0\)
thì \(x+1;x+7\)khác dấu
th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)
th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)
vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)
a) (2x - 3) = 5
<=> 2x - 3 = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
=> x = 4
b) (5x - 3) = 1/2
<=> 5x - 3 = 1/2
<=> 5x = 1/2 + 3
<=> 5x = 7/2
<=> x = 7/10
=> x = 7/10
c) (x + 1)(x + 7) < 0
<=> x = -1; -7
<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)
<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)
<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)
Vậy: -7 < x < -1
e) \(\left(x-3\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left(x-3\right)=0\) ( \(x^2+1>0\forall x\))
\(\Rightarrow x=3\)
đ) \(4.8^2=2^x\)
\(2^2.\left(2^3\right)^2=2^x\)
\(2^2.2^6=2^x\)
\(2^8=2^x\)
\(\Rightarrow x=8\)
d) \(\left|x+3\right|=8\)
\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)
mấy câu trên dễ rồi tự làm em nhé
\(a,234-\left(x-56\right)=789\)
\(\Leftrightarrow x-56=234-789\)
\(\Leftrightarrow x-56=-555\)
\(\Leftrightarrow x=\left(-555\right)+56=-499\)
Vậy x = -499
b) \(\frac{x+3}{-5}=\frac{x-15}{4}\)
\(\Leftrightarrow4\left(x+3\right)=-5\left(x-15\right)\)
\(\Leftrightarrow4x+12=-5x+75\)
\(\Leftrightarrow4x+12-\left(-5x\right)=75\)
\(\Leftrightarrow4x-\left(-5x\right)+12=75\)
\(\Leftrightarrow4x+5x=63\)
\(\Leftrightarrow9x=63\)
\(\Leftrightarrow x=7\)
Vậy x = 7
c) \(8\left(x-1\right)-7=2\left(x+2\right)+5\)
\(\Leftrightarrow8x-8-7=2x+4+5\)
\(\Leftrightarrow8x-8-7-2x+4=5\)
\(\Leftrightarrow8x-2x-8-7+4=5\)
\(\Leftrightarrow8x-2x=5-4+7+8\)
\(\Leftrightarrow4x=16\)
\(\Leftrightarrow x=4\)
Vậy x = 4
d) Đặt \(D=\frac{2x+3}{x-1}=\frac{2x-2+5}{x-1}=\frac{2\left(x-1\right)+5}{x-1}=2+\frac{5}{x-1}\)
=> \(5⋮x-1\)
=> \(x-1\inƯ\left(5\right)\)
=> \(x-1\in\left\{\pm1;\pm5\right\}\)
=> \(x\in\left\{2;0;6;-4\right\}\)
a)\(\frac{1}{4}x+2x=\frac{9}{2}\)
\(x.\left(\frac{1}{4}+2\right)=\frac{9}{2}\)
\(x.\frac{9}{4}=\frac{9}{2}\)
x\(=\frac{9}{2}:\frac{9}{4}\)
\(x=2\)
b)x=2 và x =(-1)
c)x=4
a. 25%x +2x=4,5
0,25x +2x =4,5
(0,25+2)x=4,5
2,25 x=4,5
x=4,5:2,25
x=2
Bài 2:
a) \(\left(x-3\right)^3+27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0-27\)
\(\Leftrightarrow\left(x-3\right)^3=-27\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=\left(-3\right)+3\)
\(\Leftrightarrow x=0\)
b) \(-125-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3=-125-0\)
\(\Leftrightarrow\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=\left(-5\right)-1\)
\(\Leftrightarrow x=-6\)
c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}:2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d) \(2^x+2^{x+1}=24\)
\(\Leftrightarrow2^x+2^x.2=24\)
\(\Leftrightarrow2^x\left(1+2\right)=24\)
\(\Leftrightarrow2^x.3=24\)
\(\Leftrightarrow2^x=24:3\)
\(\Leftrightarrow2^x=8\)
\(\Leftrightarrow2^x=2^3\)
\(\Rightarrow x=3\)
e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
g) \(\left|x-3\right|+2x=10\)
\(\Leftrightarrow\left|x-3\right|=10-2x\)
\(\Leftrightarrow\left|x-3\right|=2.5-2x\)
\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)
(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)
Bài 1:
a) \(2^7+2^9⋮10\)
Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)
\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)
\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)
\(\Leftrightarrow\overline{C8}+\overline{D2}\)
\(\Leftrightarrow\overline{E0}\)
Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)
b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)
Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)
\(\Leftrightarrow2^{72}.5^{20}\)
Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)
c) \(3^{10}+3^{12}⋮30\)
Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)
\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)
\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)
\(\Leftrightarrow\overline{C9}+\overline{B1}\)
\(\Leftrightarrow\overline{D0}⋮10\)
(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)
a: \(\left(2x-\dfrac{1}{5}\right)^2+\dfrac{16}{25}=\left(-2022\right)^0\)
=>\(\left(2x-\dfrac{1}{5}\right)^2+\dfrac{16}{25}=1\)
=>\(\left(2x-\dfrac{1}{5}\right)^2=1-\dfrac{16}{25}=\dfrac{9}{25}\)
=>\(\left[{}\begin{matrix}2x-\dfrac{1}{5}=\dfrac{3}{5}\\2x-\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{4}{5}\\2x=-\dfrac{2}{5}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
b: Sửa đề: \(\dfrac{x-1}{2024}+\dfrac{x-2}{2023}+\dfrac{x-3}{2022}+\dfrac{x-4}{2021}=4\)
=>\(\left(\dfrac{x-1}{2024}-1\right)+\left(\dfrac{x-2}{2023}-1\right)+\left(\dfrac{x-3}{2022}-1\right)+\left(\dfrac{x-4}{2021}-1\right)=4-4=0\)
=>\(\dfrac{x-2025}{2024}+\dfrac{x-2025}{2023}+\dfrac{x-2025}{2022}+\dfrac{x-2025}{2021}=0\)
=>x-2025=0
=>x=2025
Câu a:
\(\left(\left(\right. 2 x - \frac{1}{5} \left.\right)\right)^{2} + \frac{16}{25} = \left(\right. - 2022 \left.\right)^{0}\)
Bước 1: Ta biết rằng:
\(\left(\right.-2022\left.\right)^0=1(\text{v}\overset{ˋ}{\imath}\text{ m}ọ\text{i s}\overset{ˊ}{\hat{\text{o}}}\text{ kh}\overset{ˊ}{\text{a}}\text{c 0 m}\overset{\sim}{\text{u}}\text{ 0 b}\overset{ˋ}{\overset{}{ă}}\text{ng 1})\)
Vậy ta có phương trình:
\(\left(\left(\right. 2 x - \frac{1}{5} \left.\right)\right)^{2} + \frac{16}{25} = 1\)
Bước 2: Chuyển vế:
\(\left(\left(\right. 2 x - \frac{1}{5} \left.\right)\right)^{2} = 1 - \frac{16}{25} = \frac{9}{25}\)
Bước 3: Lấy căn hai vế:
\(2 x - \frac{1}{5} = \pm \frac{3}{5}\)
TH1:
\(2 x - \frac{1}{5} = \frac{3}{5} \Rightarrow 2 x = \frac{4}{5} \Rightarrow x = \frac{2}{5}\)
TH2:
\(2 x - \frac{1}{5} = - \frac{3}{5} \Rightarrow 2 x = - \frac{2}{5} \Rightarrow x = - \frac{1}{5}\)
✅ Đáp số câu a: \(x = \frac{2}{5}\) hoặc \(x = - \frac{1}{5}\)
Câu b:
\(\frac{x - 1}{2024} + \frac{x - 2}{2023} + \frac{x - 3}{2022} + \frac{x - 4}{2021}\)
Đây không phải phương trình (không có dấu bằng), nên mình hiểu bạn đang yêu cầu rút gọn biểu thức này.
Tách từng tử riêng ra:
\(= \frac{x}{2024} - \frac{1}{2024} + \frac{x}{2023} - \frac{2}{2023} + \frac{x}{2022} - \frac{3}{2022} + \frac{x}{2021} - \frac{4}{2021}\)
Gom nhóm các phần x lại:
\(= x \left(\right. \frac{1}{2024} + \frac{1}{2023} + \frac{1}{2022} + \frac{1}{2021} \left.\right) - \left(\right. \frac{1}{2024} + \frac{2}{2023} + \frac{3}{2022} + \frac{4}{2021} \left.\right)\)
✅ Vậy biểu thức rút gọn là:
\(x \left(\right. \frac{1}{2024} + \frac{1}{2023} + \frac{1}{2022} + \frac{1}{2021} \left.\right) - \left(\right. \frac{1}{2024} + \frac{2}{2023} + \frac{3}{2022} + \frac{4}{2021} \left.\right)\)