Tìm x:
( 7x - 27 : 23 ) : 5 + 7 . 3 = 29
Help me
tìm x thuộc Z biết :
1/ 5 - | x - 4 | = -7 + [ 23 - 54 ] - [-54 +23 ]
2/ | x + 7 | = 15 + | -27 | - [ 13 -17 + |- 20 |]
3/ | x - 7 | + 24 = 24 - x + [ - 24 ] + x + | -40 |
help me
làm ơn ai đó làm bài nay hộ minh cái
1, 7x+6(3-x)=27-20+73
2,6x-5(x-7)=(27-514)-486-73)
3,2x-6=3(x-5)+25+27-15-17
4, 4(x-5)-3(x+5)=2579-2578
5,/x-5/-7(x+4)=5-7x
6,3/x+4/-2(x-1)=7-2x
7, 2/x-6/+7x-2=/x-6/+7x
8,/x+8/+5(x-2)=5x-10:2
9, /x+2/-6(x-4)=20-6x
1: Ta có: 7x+6(3-x)=27-20+73
\(\Leftrightarrow7x+18-6x=80\)
\(\Leftrightarrow x=80-18=62\)
Vậy: x=62
2: Ta có: \(6x-5\left(x-7\right)=\left(27-514\right)-486-73\)
\(\Leftrightarrow6x-5x+35=27-514-486-73\)
\(\Leftrightarrow x+35=-1046\)
\(\Leftrightarrow x=-1081\)
Vậy: x=-1081
1) Tìm x
x - 35% . x = 1/25
2) Tính
a) 17 2/31 - ( 15/17 + 6 2/31 )
b) 27 51/59 - ( 25 51/29 - 1 1/3 )
c) ( 4 5/23 - 2 2/5 + 7 7/13 ) - ( 3 5/23 - 6 6/13 )
d) 8 1/3 + 7,8 + 5 2/3 - 1,8
Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
Bài 1:
1)-7/9x2 3/4
2)2/3+1/3x-2/5
3) 3/4 x 15 1/3 - 3/4 x 43 1/3
4) (-49,1)x 13/27 - 58,9 x 13/27
5) 0,375 : (-4,5)
6) 3 1/7 :( -1 3/7)
7) 9 1/3 :4 2/3 - 2
8) ( 7 3/4 : 0,3125 + 4,5 x2 2/45) : (-8,5)
Bài 2
A= (-6/11) x 7/10 x (11/-6) x(-20)
B= (-1/9) x (-17/29) x 58/51
C=(-3/7) x 5/11 +(-5/14)x5/11
D=(1 1/27 x 12/23 x 9/14 ) :(-3/23)
Bài 3 tìm x
a) 3/7x -2/5 x+-17/33
b)(3/4 x - 9/16).(1/3 +-3/5 : x)=0
c)(x+3/5 ) . (x+1)<0
(Hạn thứ 3 nha , giúp mk vs)
Giải các phương trình sau
a ( 3x-1)^2 - (x+3)^2
b x^3-x/49 = 0
c x^2 -7x+12
d 4x^2 -3x -1 =0
e . 29-x/21 + 27-x/23 + 25-x/25 + 23-x/28 + 21-x/29
a) \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)
\(=>\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(=>\left(4x+2\right)\left(2x-4\right)=0\)
\(=>4\left(2x+1\right)\left(x-2\right)=0\)
\(=>\orbr{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=-\frac{1}{2}\\x=2\end{cases}}\)
b)\(x^3-\frac{x}{49}=0=>x\left(x^2-\frac{1}{49}\right)=0=>x\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)
\(=>x=0\)hoặc \(x=\frac{1}{7}\) hoặc \(x=-\frac{1}{7}\)
a)\(\(\left(3x-1\right)^2-\left(x+3\right)^2=0\)\)
\(\(\Leftrightarrow\left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\)\)
\(\(\Leftrightarrow\left(2x-4\right)\left(4x+2\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}2x-4=0\\4x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}}\)\)
b)\(\(x^3-\frac{x}{49}=0\)\)
\(\(\Leftrightarrow\frac{49x^3-x}{49}=0\)\)
\(\(\Leftrightarrow x\left(49x^2-1\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\49x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(7x-1\right)\left(7x+1\right)=0\end{cases}}}\)\)\
\(\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7};x=-\frac{1}{7}\end{cases}}\)\)
c)\(\(x^2-7x+12=0\)\)
\(\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)\)
d) \(\(4x^2-3x-1=0\)\)
\(\(\Leftrightarrow4x^2-4x+x-1=0\)\)
\(\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)\)
\(\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)\)
\(\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\4x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)\)
e) Tham khảo tại : [Toán 8]Giải phương trình | Cộng đồng học sinh Việt Nam - HOCMAI Forum
https://diendan.hocmai.vn/threads/toan-8-giai-phuong-trinh.290061/
_Y nguyệt_
a thiếu đề bạn nhé
b) \(x^3-\frac{x}{49}=0\)
\(\Rightarrow x\left(x^2-\frac{1}{49}\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x^2-\frac{1}{49}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}}\)
Vậy.........
c) \(x^2-7x++12=0\)
\(\Rightarrow\left(x-3,5\right)^2-0,5^2=0\)
\(\Rightarrow\left(x-3,5+0,5\right)\left(x-3,5-0,5\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}}\)
Vậy.....
d) \(4x^2-3x-1=0\)
\(\Rightarrow4x^2-3x+0,5625-1,5625=0\)
\(\Rightarrow\left(2x-0,75\right)^2-1,25^2=0\)
\(\Rightarrow\left(2x-0,75+1,25\right)\left(2x-0,75-1,25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+0,5=0\\2x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-0,25\\x=1\end{cases}}}\)
Vậy.....
\(\dfrac{29-x}{21}\)+\(\dfrac{27-x}{23}\)+\(\dfrac{25-x}{25}\)+\(\dfrac{23-x}{27}\)+\(\dfrac{21-x}{29}\)=\(\dfrac{(29-x+1}{21}\)+\(\dfrac{(27-x+1)}{23}\)+\(\dfrac{(25-x+1)}{25}\)+\(\dfrac{(23-x+1)}{21}\)=-5 +5
GIẢI nốt hộ mình với ạ
Tìm x biết:a) 29-x/21 + 27-x/23 + 25-x/25 + 23-x/27 + 21-x/29
b)x-10/30 + x-14/43 + x-5/95 + x-148/8 = 0
a, bổ sung đề
\(\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\ne0\right)=0\Leftrightarrow x=50\)
ai giải cụ thể giùm em với
Giải phương trình sau
a) x-5/100+x-4/101+x-3/102=x-100/5+x-101/4+x-102/3
b) 29-x/21+27-x/23+25-x/25+23-x/27+21-x/29=-5