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CÂU16
a,5[-8]2[-3] b,3[-5]2+2[-5]-20 c,34[15-10]-15[34-10]
=5.2{-8.[-3]} =3. 25+{[-10]-20} =34.15-340-15. 34-150
=10. 24 =75+[-30] =0+[-340-150]
=2400 =45 =-490
d,27[-17]+[-17]73 e, 512[2-128]-128[-512]
=-17[27+73] =512. [-126]-128[-512]
=-17. 100 =-64512- [-65536]
-1700 =1024
CÂU 17
a,5-[10-x]=7 b, [4x-2][x+5]=0 c,2x-9=-8-9
5-10+x=7 ⇒4x-2 hoặc x+5=0 2x-9=-17
x=7-5+10 TH1:4x-2=0 TH2 x+5=0 2x=-17+9
x=12 4x=0+2 x=0-5 2x=-8
4x=2 x=-5 x=-8:2
x=2:4 x=-4
x=2/4 ko thỏa mãn vì x∈Z
Vậy x=-5
d,3[x-1]-27=0 e,5[3x+8]-7[2x+3]=16
3[x-1]=0+27 15x+40-14x+21=16
3[x-1]=27 15x-14x=16-21-40
[x-1]=27-3 x=-15
[x-1]=24
⇒x-1 =24 hoặc -24
TH1:x-1 =24 TH2 :x-1 =-24
x=24+1 x=-24+1
x=25 x=-23
Vậy x=25 hoặc -23
a) 5 – (10 – x) = 7
<=> 50 - 10 + x = 7
<=> 40 + x = 7
<=> x = -33
b) (4x – 2)(x + 5) = 0
<=> \(\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
c) 2x – 9 = -8 - 9
<=> 2x = -8 - 9 + 9
<=> 2x = -8
<=> x = -4
e) 5.(3x + 8) –7.(2x + 3) = 16
<=> 15x + 40 - 14x - 21 - 16 = 0
<=> x + 3 = 0
<=> x = -3
a) 5-(10-x)=7.
\(\Leftrightarrow\)5-10+x=7.
\(\Leftrightarrow\)x=7+10-5=12.
b) (4x-2)(x+5)=0.
\(\Leftrightarrow\)2(2x-1)(x+5)=0.
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x-1=0\\x+5=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
c) 2x-9=-8-9.
\(\Leftrightarrow\)2x=-8-9+9.
\(\Leftrightarrow\)2x=-8.
\(\Leftrightarrow\)x=-4.
e)5(3x+8)-7(2x+3)=16.
\(\Leftrightarrow\)15x+40-14x-21=16.
\(\Leftrightarrow\)15x-14x=16+21-40.
\(\Leftrightarrow\)x=-3.
a) (2x-5) + 17 = 6
2x - 5 = 6 - 17
2x - 5 = -11
2x = -11 + 5
2x = -6
x = -6 : 2
x = -3
* Các câu b→e bạn cũng làm tương tự theo trật tự như vậy là được
* Các câu từ g → l thì bạn áp dụng lí thuyết sau:
Tích của hai số bằng 0 khi một trong hai số đó bằng 0
VD : g) x(x+7)=0
⇒ hoặc là x = 0 hoặc là x+7 = 0
( Bạn làm phép tính nhớ bỏ dấu ngoặc vuông trước nhé )
b: \(\Leftrightarrow2\left(4-3x\right)=14\)
=>4-3x=7
=>3x=-3
=>x=-1
c: \(\Leftrightarrow3\left(7-x\right)=-18+12=-6\)
=>7-x=-2
=>x=9
d: \(\Leftrightarrow3x-2=-\dfrac{1}{8}\)
=>3x=15/8
=>x=5/8
e: \(\Leftrightarrow5\left(3x-2x\right)=-15\)
=>x=-3
g: =>x=0 hoặc x+7=0
=>x=0 hoặc x=-7
h: =>x+12=0 hoặc x-3=0
=>x=3 hoặc x=-12
k: =>x=0 hoặc x+2=0 hoặc 7-x=0
=>\(x\in\left\{0;-2;7\right\}\)
l: =>x-1=0 hoặc x+2=0 hoặc x+3=0
=>\(x\in\left\{1;-2;-3\right\}\)
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) x - 2 = -6
x = -6 + 2
x = -4
2) -5 . x - ( -3 ) =13
-5 . x = 13 + ( -3 )
-5 . x = 10
x = 10 : ( -5 )
x = -2
\(a)5-\left(10-x\right)=7\\ \Leftrightarrow5-10+x=7\\ \Leftrightarrow x=7-5+10\\ \Leftrightarrow x=12\)
\(b)\left(4x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\)\(4x-2\) hoặc \(x-5\) bằng 0
TH1: \(\left(4x-2\right)\left(x-5\right)=0\\ \Leftrightarrow4x-2=\frac{0}{x-5}\\ \Leftrightarrow4x=0+2\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\frac{2}{4}=\frac{1}{2}\)
TH2: \(\left(4x-2\right)\left(x-5\right)=0\\ \Leftrightarrow x-5=\frac{0}{4x-2}\\ \Leftrightarrow x=0+5\\ \Leftrightarrow x=5\)
Vậy \(x=\frac{1}{2}\) hoặc \(x=5\)
\(c)2x-9=-8-9\\ \Leftrightarrow2x=-8-9+9\\ \Leftrightarrow2x=-8-\left(9-9\right)\\ \Leftrightarrow2x=-8\\ \Leftrightarrow x=-\frac{8}{2}\\ \Leftrightarrow x=-4\)
\(d)3\cdot\left|x-1\right|-27=0\\ \Leftrightarrow3\cdot\left|x-1\right|=0+27\\ \Leftrightarrow3\cdot\left|x-1\right|=27\\ \Leftrightarrow\left|x-1\right|=\frac{27}{3}\\ \Leftrightarrow\left|x-1\right|=9\\ \Leftrightarrow x-1=\pm9\)
TH1; \(x-1=9\\ \Leftrightarrow x=9+1=10\)
TH2: \(x-1=-9\\ \Leftrightarrow x=-9+1=8\)
\(e)5\cdot\left(3x+8\right)-7\left(2x+3\right)=16\\ \Leftrightarrow\left(15x+40\right)-\left(14x+21\right)=16\\ \Leftrightarrow15x+40-14x-21=16\\ \Leftrightarrow\left(15x-14x\right)+\left(40-21\right)=16\\ \Leftrightarrow x+19=16\\ \Leftrightarrow x=16-19=-3\)