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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, 12 - (2\(x^2\) - 3) = 7
2\(x^2\) - 3 = 12 - 7
2\(x^2\) - 3 = 5
2\(x^2\) = 8
\(x^2\) = 4
\(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
a) ( 4x - 2 )x + 5 = 0
=> \(\hept{\begin{cases}4x-2=0\\x+5=0\end{cases}}\)
+ TH1 : 4x - 2 = 0
4x = 2
x = 0,5 ( vô lí nếu x c Z )
+ TH2 : x + 5 = 0
x = -5
=> x c \(\hept{ }\)0,5 ; -5 ( đóng ngoặc nhọn )
b) 2x - 9 = -8 -9
2x - 9 = -17
2x = -8
x = -4
c) 5( 3x + 8 ) - 7 . ( 2x + 3 ) = 16
15x + 40 - ( 14x + 21 ) = 16
15x + 40 - 14x - 24 = 16
( 15x -14x ) + (40 - 24 ) = 16
x + 16 = 16
x = 0
Xin lỗi phần c mk hơi sai sót !!! Mk sẽ làm lại phần c :
c) 5( 3x + 8 ) - 7( 2x + 3 ) = 16
15x + 40 - ( 14x + 21 ) = 16
15x + 40 - 14x - 21 = 16
( 15x - 14x ) + ( 40 - 21 ) = 16
x + 19 = 16
=> x = -3
a,x.(x+7)=0
suy ra x=o hoặc x+7=0
vs x+7=0
x=0+7
x=7
vậy x=0 hoặc x=7
b(2+2x)(7-x)=0
suy ra 2+2x=0 hoặc 7-x=0
vs2+2x=0 vs7-x=0
2x =0-2 x=0+7
2x =(-2) x=7
x=(-2);2
x=-1
vậy x=-1 hoặc x=7
d(x^2-9)(3x+15)=0
suy ra x^2-9=0 hoặc 3x+15=0
vsx^2-9=0 vs 3x+15=0
x^2 =0+9 3x =0-15
x^2 =9 3x =-15
x^2 =3^2 x=(-15):3
suy ra x=3 hoặc x=-3 x=-5
vậy x=3 x=-3 hoặc x=-5
e,(4x-8)(x^2+1)=0
suy ra4x-8=0 hoặc x^2+1=0
vs 4x-8=0 vs x^2+1=0
4x =0+8 x^2 =0-1
4x =8 x^2 =-1
x =8:4 x^2 =-1^2 hoặc 1^2
x =2 suy ra x=-1 hoặc x=1
vậy x=2, x=-1 hoặc x=1
1)=>3(x-5)(2x+9)+3(x-5)=0=>(x-5)(6x+30)
=>x-5=0=>x=5
6x+30=0=>x=-5
2)=>x^2-16=0=>x=+-4
12-4x=0=>x=3
3)=>9-x^2=0=>x=+-3
4x-8=0=>x=2
4)=>8-x^3=0=>x=3
5^x-125=0=>x=2
5)=>2^x.2^x=8=>2^2x=8=>2x=3=>x=1,5
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
a) (4x – 2)(x + 5) = 0
⇔ \(\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}4x=2\\x=-5\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
Vậy ...
b) 2x – 9 = -8 – 9
⇔ 2x = - 8
⇔ x = - 4
Vậy x = - 4
c) 3.| x -1 | - 27 = 0
⇔ 3 . | x - 1| = 27
⇔ | x - 1| = 9
⇔ \(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
Vậy x ∈ { 10 ; - 8}
d) 5.(3x + 8) –7.(2x + 3) = 16
⇔ 15x + 40 - 14x - 21 = 16
⇔ x + 19 = 16
⇔ x = - 3
Vậy ..
a)\(\left(4x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
Vậy...
b)\(2x-9=-8-9\\ \Leftrightarrow2x-9=-17\\ \Leftrightarrow2x=-8\\ \Leftrightarrow x=-4\)
Vậy...
c)\(2\left|x-1\right|-27=0\\ \Leftrightarrow\left|x-1\right|=\frac{27}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{27}{2}\\x-1=-\frac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{2}\\x=-\frac{25}{2}\end{matrix}\right.\)
Vậy...