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a)
-2x+3=-7
-2x=-7-3
-2x=-10
x=-10:-2
x=5
b)(-3)x+1=-8
-3x=-8-1
-3x=-9
x=-9 :-3
x=3
c)[2x+1]=5 ( cái này à trị tuyệt đối đúng k ?) nếu dấu [ là dấu giá trị tuyệt đối
=>\(\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
=>\(\orbr{\begin{cases}2x=4\\2x=-6\end{cases}}\)
=>\(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
vậy x \(\in\left\{2;-3\right\}\)
d)|2x-1|-3=18
|2x-1|=21
=> \(\orbr{\begin{cases}2x-1=21\\2x-1=-21\end{cases}}\)
=>\(\orbr{\begin{cases}2x=22\\2x=-20\end{cases}}\)
=>\(\orbr{\begin{cases}x=11\\x=-10\end{cases}}\)
vậy \(x\in\left\{11;-10\right\}\)
Bài 1 Tìm x biết:
a)65-(29-x)=32
65 -29+x=31
x=31-65+29
x=-5
b)(x+5)-(x+23)=x-34
x+5 -x +23 = x-34
(x-x)+ (23+5)=x-34
0+28=x-34
28=x-34
28+34=x
62=x
=>x=62
c)(16-x)+(x-38)=x+44
16-x+x-38=x+44
-x+x-x=44-16+38
-x=36
=>x=-36
d)-12+3(-x+7)=-18
3(-x+7)=-18+12
3(-x+7)=-6
-x+7=-6:3
-x+7=-2
-x=-2-7
-x=-9
=>x=9
Baif 2
d)|7-x|=10
=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)
e)(x-6).(7-2x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)
f)(9-x).(2x+8)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
g)x(-x+8).(-3x-18)=0
\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)
h)(-x+8).(x-54).(-24-x)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)
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.
1e) Để \(\frac{2x-1}{x-3}\) nguyên thì \(2x-1⋮x-3\)
\(\Leftrightarrow2x-6+5⋮x-3\)
\(\Leftrightarrow2\left(x-3\right)+5⋮x-3\)
Do \(2\left(x-3\right)⋮x-3\) \(\Rightarrow5⋮x-3\)
\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)
Vậy:...................
\(12\left(x-3\right)=5\left(x-1\right)+4\)
\(\Rightarrow12x-36=5x-5+4\)
\(\Rightarrow12x-5x=-5+4+36\)
\(\Rightarrow7x=35\)
\(\Rightarrow x=5\)
\(-6\left(x-7\right)+2\left(x+10\right)=x-18\)
\(-6x+42+2x+20=x-18\)
\(-6x+2x-x=-18-20-42\)
\(-5x=-80\)
\(x=16\)
\(7\left(2x-1\right)-12\left(x-5\right)=3\)
\(\Rightarrow14x-7-12x+60=3\)
\(\Rightarrow14x-12x=3-60+7\)
\(\Rightarrow2x=-50\)
\(\Rightarrow x=-25\)
\(\left|2x-1\right|=17\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=17\\2x-1=-17\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-8\end{cases}}\)
\(\left|8-3x\right|=14\)
\(\Leftrightarrow\orbr{\begin{cases}8-3x=14\\8-3x=-14\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{22}{3}\end{cases}}\)
\(\left(2x-1\right)^8=\left(2x-1\right)^{18}\)
Ta thừa nhận kết luận sau: Số 1 với bất kì số mũ nào cũng là chính nó.
\(\Rightarrow2x-1=1\) thì \(\left(2x-1\right)^8=\left(2x-1\right)^{18}\)
Giải \(2x-1=1\) ta có:
\(2x-1=1\Leftrightarrow2x=2\Leftrightarrow x=1\)
tthctv xem lại nhá :)
\(\left(2x-1\right)^8=\left(2x-1\right)^{18}\)
\(\Leftrightarrow\)\(\left(2x-1\right)^{18}-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)^8\left[\left(2x-1\right)^{10}-1\right]=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)\left(2x-1-1\right)\left(2x-1+1\right)=0\)
\(\Leftrightarrow\)\(2x\left(2x-1\right)\left(2x-2\right)=0\)
\(\Leftrightarrow\)\(2x=0\)\(\Leftrightarrow\)\(x=0\)
Hoặc \(2x-1=0\)\(\Leftrightarrow\)\(x=\frac{1}{2}\)
Hoặc \(2x-2=0\)\(\Leftrightarrow\)\(x=1\)
Vậy \(x=0\)\(x=\frac{1}{2}\) hoặc \(x=1\)
Chúc bạn học tốt ~