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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:...................
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. Tìm số nguyên x biết
a.3x+27=9
\(\Leftrightarrow3x=-18\)
\(\Leftrightarrow x=-6\)
b)\(2x^2-1=49\)
\(\Leftrightarrow2x^2=50\)
\(\Leftrightarrow2x^2=2.5^2\)
\(\Leftrightarrow x=\pm5\)
c)2x+12=3(x-7)
\(\Leftrightarrow2x+12=3x-21\)
\(\Leftrightarrow33=x\) hay \(x=33\)
d)\(\left|-9-x\right|-5=12\)
\(\Leftrightarrow\left|-9-x\right|=17\)
\(\Leftrightarrow\left[{}\begin{matrix}-9-x=17\\-9-x=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-26\\x=8\end{matrix}\right.\)
đ)(x-5)(x+6)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
e)(3-x)(x+7)=0
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
2.cho biểu thức
A=(-a-b+c)-(-a-b-c)
a)rút gọn
\(A=\left(-a+b+c\right)-\left(-a-b-c\right)\)
\(A=-a+b+c+a+b+c\)
\(A=2c\)
b)Tính giá trị của A khi a=1;b=-1;c=-2
Thay a=1;b=-1;c=-2 vào A ta có
\(A=2c\)
\(A=2.\left(-2\right)\)
\(A=-4\)
a2 và 2a
Nếu a=0 thì 02=2.0
Nếu a=1 thì 12=2.1
Nếu a=2 thì 22=2.2
Nếu a>2 hoặc a<0 thì a2=a.a=a+a+...+a(a số a) ; 2a=2.a=a+a
=>a2>2.a
Đáp án cần chọn là: D