a.    b, 315 – (135 – x) = 215 ;  <...">
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6 tháng 3 2020

a) |2x + 1| - 19 = -7

=> \(\left|2x+1\right|=-7+19=12\)

=> \(\left[{}\begin{matrix}2x+1=12\\2x+1=-12\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x=12-1=11\\2x=-12-1=-13\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{11}{2}\\x=-\frac{13}{2}\end{matrix}\right.\)

Vậy:............

b) -28 – 7. |- 3x + 15| = -70

=> \(\text{7. |- 3x + 15| = -28 - (-70) = -28 + 70 = 42}\)

=> \(\left|-3x+15\right|=42:7=6\)

=> \(\left[{}\begin{matrix}-3x+15=6\\-3x+15=-6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}-3x=6-15=-9\\-3x=-6-15=-6+\left(-15\right)=-21\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-9:\left(-3\right)=3\\x=x=-21:\left(-3\right)=7\end{matrix}\right.\)

Vậy:.....................

c) |18 – 2. |-x + 5|| = 12

=> \(\left[{}\begin{matrix}18-2.\left|-x+5\right|=12\\18-2.\left|-x+5\right|=-12\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2.\left|-x+5\right|=18-12=6\\2.\left|-x+5\right|=18-\left(-12\right)=18+12=30\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left|-x+5\right|=6:2=3\\\left|-x+5\right|=30:2=15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}-x+5=3\\-x+5=-3\end{matrix}\right.\\\left[{}\begin{matrix}-x+5=15\\-x+5=-15\end{matrix}\right.\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}-x=3-5=-2\\-x=-3-5=-8\end{matrix}\right.\\\left[{}\begin{matrix}-x=15-5=10\\-x=-15-5=-20\end{matrix}\right.\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\\\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\end{matrix}\right.\)

7 tháng 3 2020

thank bn ✰❤❤✔

5 tháng 3 2020

1/ (a – b + c) – (a + c) = -b

a-b+c-a-c=-b

-b=-b

2/ (a + b) – (b – a) + c = 2a + c

a+b-b+a+c=2a+c

2a+c=2a+c

3/ - (a + b – c) + (a – b – c) = -2b

-a-b+c+a-b-c=-2b

-(b.2)=-2b

-2b=-2b

4/ a(b + c) – a(b + d) = a(c – d)

ab+ac-ab+ad=a(c-d)

ac-ad=a(c-d)

a(c-d)=a(c-d)

5/ a(b – c) + a(d + c) = a(b + d)

ab-ac+ad+ac=a(b+d)

ab+ad=a(b+d)

a(b+d)=a(b+d)

6/ a.(b – c) – a.(b + d) = -a.( c + d)

ab-ac-ab=ad=-a(c+d)

-ac+ad=-a(c+d)

-a(c+d)=-a(c+d)

5 tháng 3 2020

thank bn

2 tháng 8 2017

a, x : [(1800 + 600) : 30] = 560 : (315 - 35)
<=> x : (2400 : 30) = 560 : 280
<=> x : 80 = 2
<=> x = 160
b, x - 6 : 2 - (48 - 24) : 2 : 6 - 3 = 0
<=> x - 3 - 24 : 2 : 6 - 3 = 0
<=> x - 3 - 2 - 3 = 0
<=> x - 8 = 0
<=> x = 8
c, 390 - (x - 7) = 169 : 13
<=> 390 - (x - 7) = 13
<=> x - 7 = 377
<=> x = 384
d, (x - 140) : 7 = 33 - 23.3
<=> (x - 140) : 7 = 33 - 24
<=> (x - 140) : 7 = 3
<=> x - 140 = 21
<=> x = 161
@Đỗ Thị Huyền Trang

2 tháng 8 2017

a) x = 160

b) x= 8

c) x = 384

d) x = 539

16 tháng 5 2018

a) \(\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)\cdot\frac{11}{7}=\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{7}\div\frac{11}{7}\)

\(\left(\frac{9}{2}-2x\right)=1\)

\(2x=\frac{9}{2}-1\)

\(x=\frac{7}{2}\div2\)

\(x=\frac{7}{4}\)

b) \(|\frac{3}{4}\cdot x-\frac{1}{2}|-1=\frac{1}{4}\)

\(|\frac{3}{4}\cdot x-\frac{1}{2}|=\frac{1}{4}+1\)

\(|\frac{3}{4}\cdot x|=\frac{5}{4}+\frac{1}{2}\)

\(x=\frac{7}{4}\div\frac{3}{4}\)

\(x=\frac{7}{3}\)

c) \(\frac{1}{4}-|3-x|=-\frac{3}{4}\)

\(|3-x|=\frac{1}{4}-\left(-\frac{3}{4}\right)\)

\(|3-x|=1\)

\(x=3-1\)

\(\Rightarrow x=2\)

d) \(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=1,4\)

\(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=\frac{7}{5}\)

\(4\cdot\left(x-\frac{6}{7}\right)=\frac{7}{5}+\frac{3}{5}\)

\(4\cdot\left(x-\frac{6}{7}\right)=2\)

\(\left(x-\frac{6}{7}\right)=2\div4\)

\(x=\frac{1}{2}+\frac{6}{7}\)

\(x=\frac{19}{14}\)

\(\)

16 tháng 5 2018

k cho mình nhé 

chúc bạn học giỏi

a) Ta có:

\(\frac{3}{x+2}=\frac{5}{2x+1}\)

\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)

\(\Rightarrow6x+3=5x+10\)

\(\Rightarrow6x-5x=10-3\)

\(\Rightarrow x=7\)

b)Ta có:

\(\frac{5}{8x-2}=\frac{-4}{7-x}\)

\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)

\(\Rightarrow35-5x=-32x+8\)

\(\Rightarrow-5x+32x=8-35\)

\(\Rightarrow27x=-27\)

\(\Rightarrow x=-1\)

c) Ta có:

\(\frac{4}{3}=\frac{2x-1}{x}\)

\(\Rightarrow4x=3\left(2x-1\right)\)

\(\Rightarrow4x=6x-3\)

\(\Rightarrow3=6x-4x=2x\)

\(\Rightarrow x=\frac{3}{2}\)

d)Ta có:

\(\frac{2x-1}{3}=\frac{3x+1}{4}\)

\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)

\(\Rightarrow8x-4=9x+3\)

\(\Rightarrow8x-9x=3+4\)

\(\Rightarrow-x=7\Rightarrow x=-7\)

e)Ta có:

\(\frac{4}{x+2}=\frac{7}{3x+1}\)

\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)

\(\Rightarrow12x+4=7x+14\)

\(\Rightarrow12x-7x=14-4\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

f)Ta có:

\(\frac{-3}{x+1}=\frac{4}{2-2x}\)

\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)

\(\Rightarrow-6+6x=4x+4\)

\(\Rightarrow6x-4x=4+6\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

9 tháng 8 2016

Bạn ơi ; tách từng bài ra cho dễ làm :

1.7C-C= 7^2016-7

  C  = ( 7^2016-7 ) :6

\(C=7+7^2+7^3+.....+7^{2016}\)

\(\Rightarrow7C=7^2+7^3+7^4+...+7^{2017}\)

\(\Rightarrow7C-C=\left(7^2+7^3+.....+7^{2017}\right)-\left(7+7^2+7^3+....+7^{2016}\right)\)

\(\Rightarrow6C=2^{2017}-7\)

\(\Rightarrow C=\frac{2^{2017}-7}{6}\)

a.Ta có:|2x-1|=2x-1\(\Leftrightarrow\)2x-1\(\ge\)0\(\Leftrightarrow\)x\(\ge\)\(\dfrac{1}{2}\)

|2x-1|=1-2x\(\Leftrightarrow\)2x-1<0\(\Leftrightarrow\)x<\(\dfrac{1}{2}\)

ĐK:\(x\ge\dfrac{1}{2}\)

\(2x-1=2x-1\)

\(\Leftrightarrow2x-1-2x+1=0\)

\(\Leftrightarrow0x=0\)

\(\Rightarrow\)Tập no của PT là S={\(\forall x\)|x\(\ge\dfrac{1}{2}\)}

b.|0,5-3x|=3x-0,5\(\Leftrightarrow\)x<2,5

=0,5-3x\(\Leftrightarrow x\ge2,5\)

ĐK:x<2,5

Gỉai

0,5-3x=3x-0,5

\(\Leftrightarrow\)0,5-3x-3x+0,5=0

\(\Leftrightarrow\)1-6x=0

\(\Leftrightarrow x=\dfrac{1}{6}\)(TMĐKXĐ)

\(\Rightarrow\)tập no của PT là S={\(\dfrac{1}{6}\)}

c.|5x+1-10x|=0,5\(\Leftrightarrow\)|1-5x|=0,5\(\Leftrightarrow x< \dfrac{1}{5}\)

\(\Leftrightarrow\)|1-5x|=-0,5\(\Leftrightarrow\)x\(\ge\dfrac{1}{5}\)

ĐK:\(x< \dfrac{1}{5}\)

Gỉai

1-5x=0,5

\(\Leftrightarrow5x=0,5\)

\(\Leftrightarrow x=0,1\)(loại)

\(\Rightarrow pt\) trên vô nghiệm

d.|x+2|-|x-7|=0

ĐK:x\(\ne\pm2\);x\(\ne\pm7\)

Gỉai

\(\left\{{}\begin{matrix}x+2-x+7=0\\x-2-x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-9=0\\-2x-9=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-9=0\left(KTMĐKXĐ\right)\\x=-4,5\left(TMĐKXĐ\right)\end{matrix}\right.\)

\(\Rightarrow\)tập no của phương trình là S={-4,5}