tìm x,y
a,x+7/12=3/4 b,4/5x= -8/35 c,y-5/7= -19/21 d,1/3+2/3.x=1/4 e, x : 3/4+1/4=_2/3 f,2x-25%x=1/2
g,1/3x+2/5(x-1)=0
Bài 1 tìm x
l) (x + 9) . (x2 – 25) = 0
e) |x - 4 |< 7
f) 40 < 31 + |x |< 47
g) | x + 3| ≤ 2
m) (-5x + 20).(x3 – 8) = 0
a) (x + 1).(y - 2) = 5
b) (x - 5).(y + 4) = -7
c) (x + 1)2 + (y – 1)2 = 0
d) (2x – 18)2 + ( y + 37)2 = 0
k |x-40|+|x-y+10|_<0
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
Bài 1:
l) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-9;5;-5\right\}\)
e) Ta có: |x-4|<7
mà \(\left|x-4\right|\ge0\forall x\)
nên \(\left|x-4\right|\in\left\{0;1;2;3;4;5;6\right\}\)
\(\Leftrightarrow x-4\in\left\{0;1;-1;2;-2;3;-3;4;-4;5;-5;6;-6\right\}\)
hay \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)
Vậy: \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)
f) Ta có: \(40< 31+\left|x\right|< 47\)
\(\Leftrightarrow\left|x\right|+31\in\left\{41;42;43;44;45;46\right\}\)
\(\Leftrightarrow\left|x\right|\in\left\{10;11;12;13;14;15\right\}\)
hay \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)
Vậy: \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)
g) Ta có: \(\left|x+3\right|\le2\)
\(\Leftrightarrow\left|x+3\right|\in\left\{0;1;2\right\}\)
\(\Leftrightarrow x+3\in\left\{0;1;-1;2;-2\right\}\)
hay \(x\in\left\{-3;-2;-4;-1;-5\right\}\)
Vậy: \(x\in\left\{-3;-2;-4;-1;-5\right\}\)
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
Tìm x biết a) x(x-25)=0 b)2x(x-4)-x(2x-1)=-28 c)x^2 -5x=0 d)(x-2)^2-(x+1)(x+3)=-7 e)(3x+5).(4-3x)=0 f)x^2-1/4=0
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
Bài 1 giải các phương trình sau
A. 5x-25=0
4x-1=3x-2
B. 3/4-3x=0
C. 3x-2=2x+3
(2x-3)(x+3)=3x+9
D. 2(x-3)=5(x+4)
E. 8x-3/5=2x+8/8
X-5x+2/6=7-3x/4
G. 7x-3/5=5x+7/7
H. (3x-5)(7x+5)=0
L. (1/2-3/43/4)(5-2x)=0
M. (2x+7)(x-5)(5x+1)=0
M.x+1/x-3 - 1/x-1=2/(x-1)(x-3)
\(A,5x-25=0\)
\(\Leftrightarrow5x-5^2=0\)
\(\Leftrightarrow5\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Rightarrow x=1\)
Chúc bạn học tốt !
* 4x - 1 = 3x - 2
⇔ 4x - 3x = -2 + 1
⇔ x = -1
Vậy tập nghiệm của pt là S = {-1}
* \(\frac{3}{4}-3x=0\)
⇔ \(\frac{3}{4}-\frac{3x.4}{4}=0\)
⇒ 3 - 12x = 0
⇔ 12x = 3
⇔ x = \(\frac{3}{12}=\frac{1}{4}\)
Vậy tập nghiệm của pt là S = \(\left\{\frac{1}{4}\right\}\)
* 3x - 2 = 2x + 3
⇔ 3x - 2x = 3 + 2
⇔ x = 5
Vậy tập nghiệm của pt là S = {5}
* 2(x - 3) = 5(x + 4)
⇔ 2x - 6 = 5x + 20
⇔ 2x - 5x = 20 + 6
⇔ -3x = 26
⇔ x = \(\frac{-26}{3}\)
Vậy tập nghiệm của pt là S = \(\left\{\frac{-26}{3}\right\}\)
Câu 3. Giải các phương trình sau bằng cách đưa về dạng ax+b= 0
1. a, 3x-2=2x-3; b, 3-4y+24+6y=y+27+3y
c, 7-2x=22-3x; d, 8x-3=5x+12
e, x-12+4x=25+2x-1; f, x+2x+3x-19=3x+5
g, 11+8x-3=5x-3+x; h, 4-2x+15=9x+4-2
2. a, 5-(x-6)=4(3-2); b, 2x (x+2)2-8x2=2(x-2) (x2+2x-4)
c, 7-(2x+4)=-(x+4); d, (x-2)3+(3x-1) (3x+1)=(x+1)3
e, (x+1) (2x-3)=(2x-1) (x+5); f, (x-1)3-x(x+1)2=5x (2-x)-11 (x+2)
g, (x-1)-(2x-1)=9-x; h, (x-3) (x+4)-2(3x-2)=(x-4)2
i, x(x+3)2-3x=(x+2)3+1; j, (x+1) (x2-x+1)-2x=x(x+1) (x-1)
3. a, 1,2-(x-0,8)=-2(0,9+x); b, 3,6-0,5 (2x+1)=x-0,25 (2-4x)
c, 2,3x-2 (0,7+2x)= 3,6-1,7x; d, 0,1-2 (0,5t-0,1)=2 (t-2,5)-0,7
e, 3+2,25x+2,6= 2x+5+0,4x; f, 5x+3,48-2,35x= 5,38-2,9x+10,42
Copy có khác, ko đọc đc j!!! ʌl
Câu 3:
1)
a) Ta có: 3x−2=2x−33x−2=2x−3
⇔3x−2−2x+3=0⇔3x−2−2x+3=0
⇔x+1=0⇔x+1=0
hay x=-1
Vậy: x=-1
b) Ta có: 3−4y+24+6y=y+27+3y3−4y+24+6y=y+27+3y
⇔27+2y=27+4y⇔27+2y=27+4y
⇔27+2y−27−4y=0⇔27+2y−27−4y=0
⇔−2y=0⇔−2y=0
hay y=0
Vậy: y=0
c) Ta có: 7−2x=22−3x7−2x=22−3x
⇔7−2x−22+3x=0⇔7−2x−22+3x=0
⇔−15+x=0⇔−15+x=0
hay x=15
Vậy: x=15
d) Ta có: 8x−3=5x+128x−3=5x+12
⇔8x−3−5x−12=0⇔8x−3−5x−12=0
⇔3x−15=0⇔3x−15=0
⇔3(x−5)=0⇔3(x−5)=0
Vì 3≠0
nên x-5=0
hay x=5
Vậy: x=5
a) 3x - 2 = 2x - 3
\(\Leftrightarrow\) 3x - 2 - 2x + 3 = 0
\(\Leftrightarrow\) x + 1 = 0
\(\Rightarrow\) x = -1
b) 3 - 4y + 24 + 6y = y + 27 + 3y
\(\Leftrightarrow\) 3 - 4y + 24 + 6y - y - 27 - 3y = 0
\(\Leftrightarrow\) -2y = 0
\(\Rightarrow\) y = 0
c)7 - 2x = 22 - 3x
\(\Leftrightarrow\) 7 - 2x - 22 + 3x = 0
\(\Leftrightarrow\) -15 + x = 0
\(\Rightarrow\) x = 15
d) 8x - 3 = 5x + 12
\(\Leftrightarrow\) 8x - 3 - 5x - 12 = 0
\(\Leftrightarrow\)3x -15 = 0
\(\Leftrightarrow\) 3x = 15
\(\Rightarrow\) x = 5
e) x - 12 + 4x = 25 + 2x - 1
\(\Leftrightarrow\) x - 12 + 4x - 25 - 2x + 1 = 0
\(\Leftrightarrow\) 3x - 36 = 0
\(\Leftrightarrow\) 3x = 36
\(\Rightarrow\) x = 12
f ) x + 2x + 3x - 19 = 3x + 5
\(\Leftrightarrow\) x + 2x + 3x - 19 - 3x - 5 = 0
\(\Leftrightarrow\)3x - 24 = 0
\(\Leftrightarrow\) 3x = 24
\(\Rightarrow\) x = 8
g) 11+ 8x - 3 = 5x - 3 +x
\(\Leftrightarrow\)8x + 8 = 6x - 3
\(\Leftrightarrow\)8x - 6x = -3 - 8
\(\Leftrightarrow\)2x = -11
\(\Rightarrow\)x = \(-\frac{11}{2}\)
h) 4 - 2x +15 = 9x + 4 -2
\(\Leftrightarrow\)19 - 2x = 7x + 4
\(\Leftrightarrow\)-2x - 7x = 4 - 19
\(\Leftrightarrow\)-9x = -15
\(\Rightarrow\)x = \(\frac{15}{9}\) = \(\frac{5}{3}\)
2)
a) \(5-\left(x-6\right)=4\cdot\left(3-2\right)\)
\(\Leftrightarrow5-x+6=12-8\)
\(\Leftrightarrow11-x=4\)
\(\Rightarrow x=7\)
b) \(2x\cdot\left(x+2\right)^2-8x^2=2\cdot\left(x-2\right)\cdot\left(x^2+2x+4\right)\)
\(\Leftrightarrow2x\cdot\left(x^2+4x+4\right)-8x^2=2\cdot\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)
\(\Leftrightarrow8x+16=0\)
\(\Rightarrow x=-2\)
c) \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow-2x+x=-4-3\)
\(\Leftrightarrow-x=-7\)
\(\Rightarrow x=7\)
d) \(\left(x-2\right)^3+\left(3x-1\right)\cdot\left(3x+1\right)=\left(x+1\right)^3\)
\(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)
\(\Leftrightarrow9x-10=0\)
\(\Rightarrow x=\frac{10}{9}\)
e)\(\left(x+1\right)\cdot\left(2x-3\right)=\left(2x-1\right)\cdot\left(x+5\right)\)
\(\Leftrightarrow2x^3-3x+2x-3-2x^2-10x+x+5=0\)
\(\Leftrightarrow2-10x=0\)
\(\Rightarrow x=\frac{2}{10}=\frac{1}{5}\)
f)\(\left(x-1\right)^3-x\cdot\left(x+1\right)^2=5x\cdot\left(2-x\right)-11\cdot\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x-10x+5x^2+11x+22=0\)
\(\Leftrightarrow3x+21=0\)
\(\Rightarrow x=-7\)
g)\(\left(x-1\right)-\left(2x-1\right)=9-x\)
\(\Leftrightarrow x-1-2x+1-9+x=0\)
\(\Leftrightarrow-9=0\)
\(\Rightarrow\) Phương trình vô nghiệm
h)\(\left(x-3\right)\cdot\left(x+4\right)-2\cdot\left(3x-2\right)=\left(x-4\right)^2\)
\(\Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\)
\(\Leftrightarrow x^2-5x-8=x^2-8x+16\)
\(\Leftrightarrow x^2-5x-8-x^2+8x-16=0\)
\(\Leftrightarrow3x-24=0\)
\(\Rightarrow x=8\)
i)\(x\cdot\left(x+3\right)^2-3x=\left(x+2\right)^3+1\)
\(\Leftrightarrow x^3+6x^2+9x-3x=x^3+6x^2+12x+8+1\)
\(\Leftrightarrow x^3+6x^2+6x=x^3+6x^2+12x+9\)
\(\Leftrightarrow x^3+6x^2+6x-x^3-6x^2-12x-9=0\)
\(\Leftrightarrow-6x-9=0\)
\(\Rightarrow x=-\frac{3}{2}\)
j)\(\left(x+1\right)\cdot\left(x^2-x+1\right)-2x=x\cdot\left(x+1\right)\cdot\left(x-1\right)\)
\(\Leftrightarrow\left(x^3+1\right)-2x=x\left(x^2-1\right)\)
\(\Leftrightarrow x^3+1-2x-x^3+x=0\)
\(\Leftrightarrow1-x=0\)
\(\Rightarrow x=1\)
BT6:
A) (5x+1)^2=16/25
B) ( x - 1/3)^3=(2/3)^6
C)(8x - 1)^4=(2x+7)^4
D) (x -1)^5 =(2x-3)^5
E) ( x +1 )^8=(x+1)^10
G) (21-3)^4+(x+5)^6=0
H) 3x -4 = 81^3
a) (5x+1) ^ 2 = 4^2 : 5^ 2
( 5x+1) ^2 = (4:5) ^2
=> (5x+1) = ( 4 : 5) = 0.8
5x = 0.8 - 1
x = 0.7 : 5
x = 0,14
1. a. 4x-20=0
b. 2x+x+12=0
c. x-5=3-x
d. 7-3x=9-x
2.
a. 7+2x=22-3x
b. 8x-3=5x+12
c. x-12+4x=25+2x-1
d. x+2x+3x-19=3x+5
e. 7-(2x+4)=-(x+4)
f. (x-1)-(2x-1)=9-x
Bài 1: Tìm x, biết
a) 3/7x - 2/3x =10/21
b) 7/35 : ( x- 1/3)=2/25
c) | 2x-4 |+1=5
d) 3.| 3 - 2x | - 1=2/5
e) 3.(x-1/2)-5(x+3/5)=-x+1/5
f) (2x-1).(x+2/3)=0
g) x+4/2008+x+3/2009=x+2/2010+x+1/2011
a, \((\frac{3}{7}-\frac{2}{3})\) .x =\(\frac{10}{21}\)
\(\frac{-5}{21}\).x=\(\frac{10}{21}\)
x= -2
Mk chỉ làm 1 phần các phằn còn lại tương tự
Bài rút gọn biểu thức
a) M=|2x-3|+|x-1| với x > 1,5
b) N=|2-x|-3|x+1| với x < -1
c) P=|3x-5|+|x-2|
d) Q=|x-3|-2.|-5x|
a, \(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\Rightarrow\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\Rightarrow x=\frac{10}{21}.\left(\frac{-21}{5}\right)\Rightarrow x=-2\)
b, \(\frac{7}{35}:\left(x-\frac{1}{3}\right)=\frac{2}{25}\Rightarrow x-\frac{1}{3}=\frac{1}{5}:\frac{2}{25}\Rightarrow x=\frac{1}{10}-\frac{1}{3}=\frac{13}{30}\)
c, \(|2x-4|+1=5\)
\(\Rightarrow\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)
d, \(3.|3-2x|-1=\frac{2}{5}\)
\(\Rightarrow3.|3-2x|=\frac{7}{5}\)
\(\Rightarrow\orbr{\begin{cases}3-2x=\frac{7}{15}\\3-2x=-\frac{7}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{19}{15}\\x=\frac{26}{15}\end{cases}}\)
f, \(\left(2x-1\right).\left(x+\frac{2}{3}\right)=0\)
\(\Rightarrow\hept{\begin{cases}2x-1=0\\x+\frac{2}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{3}\end{cases}}\)
A) 2(x-3)=5(x+4)
B) 8x-3/5=2x+8/8
X-5x+2/6=7-3x/4
C) 7x-3/5=5x+7/7
D) (3x-5)(7x+5)=0
E) (½x -3/4 )(5-2x)=0
G)( 2x+7)(x-5)(5x+1)
H) x+1/x-3 - 1/x-1=2/(x-1)(x-3)