(x^2-5).(x+2)+5x=2x^2+7
Tìm x
tìm x , biết
a) 17/6- x( x-7/6)= 7/4
b) 3/35 - ( 3/5-x)= 2/7
tìm x thuộc Z , biết
3/4-5/6 < x/12 < 1 -( 2/3-1/4)
tìm x biết
a ) 2x-3=x + 1/2
b) 4x- ( x+ 1/2) = 2x - ( 1/2 - 5 )
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
Bài 3:
a) Ta có: \(2x-3=x+\dfrac{1}{2}\)
\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)
\(\Leftrightarrow x=5\)
tìm x biết
a) 5x(2x-7)+2x(8-5x)=5
b)(3x-5)(7-5x)-(5x+2)(2-3x)=4
c)5(2x-3)^2-5(x+1)^2-15(x+4)(x-4)=-10
d) 5x(x-3)(x+3)-(2x-3)^2 -5(x+2)^3+34x(x+2)=1
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
Các bạn giúp mình nhé càng nhanh càng tốt nhà
(5x-1). (2x+3)-3. (3x-1)=0
x^3 (2x-3)-x^2 (4x^2-6x+2)=0
x (x-1)-x^2+2x=5
(3x+2)(x-1)-3 (5x+2)+5 (11-4x)=25
8 (x-2)-2 (3x-4)=25
(3x+4). (5x-1)+(5x+2). (1-3x)+2=0
(5x-1). (2x+7)-(2x-3). (5x+9)
4 (x-1). (X+5)-(x+5). (X+2)=3. (X-1)(x+2)
2x^2+3 (x-1). (X+1)=5x(x+1)
4. (18-5x)-12 (3x-7)=1825. (2x-16)-6 .(x+4)
1/2x. (2/5-4x)+(2x+5).x=-13/2
Nhiều các bạn giả đùm mình nha
Thanh nhiều
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
Tìm x
1) (5x-7)(x-9)-(-x+3)(-5x+2)=2x(x-4)-(x-1)(2x+3)
2) (x-3)(-2x+5)-2x(x-4)+(4x-5)(x-3)=(x-2)(x-1)-(x^2-5x)
1, x= 2
2, x = 4
**** bạn mình trước nhé
trieu dang sai ket qua vi chua doi dau
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: \(x=-2\)
1. Rút Gọn
a) -5x (x-3).(2x+4)-(x+3)(x-3)+(5x-2)(3x+4)
b) (4x-1)x(3x+1)-5x^2x(x-3)-(x-4)x(x-5)-7(x^3-2x^2+x-1)
c) (5x-7)(x-9)-(3-x)(2-5x)-2x(x-4)
d)(5x-4)(x+5)-(x+1)(x^2-6)-5x+19
e)(9x^2-5)(x-3)-3x^2(3x+9)-(x-5)(x+4)-9x^3
g) (x-1)^2 - (x+2)^2
Thanks mn nhiều ạ
\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)
\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)
\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)
\(=-10x^3+19x^2+74x+1\)
\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)
\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)
\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)
\(=-5x^4-11x^3+24x^2+12x+7\)
\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)
\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)
\(=-2x^2-27x+57\)
\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)
\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)
\(=-x^3+4x^2+22x+5\)
\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)
\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)
\(=-9x^3-55x^2+4x+35\)
\(g,\left(x-1\right)^2-\left(x+2\right)^2\)
\(=x^2-2x+1-x^2-4x-4\)
\(=-6x-3\)
Gỉai các phương trình sau
a) 5/-x^2+5x-6 + x+3/2-x = 0
b) x/2x+2 - 2x/x^2-2x-3 = x/6-2x
c) 1/x-1 - 3x^2/x^3-1 = 2x/x^2+x+1
d) x+25/2x^2-50 - x+5/x^2-5x = 5-x/2x^2+10x
\(a,\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\left(x\ne2;x\ne3\right)\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}=0\\\Leftrightarrow\dfrac{5-\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=0 \\ \Leftrightarrow5-x^2+9=0\\ \Leftrightarrow14-x^2=0\\ \Leftrightarrow x^2=14\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{14}\\x=-\sqrt{14}\end{matrix}\right.\)
\(b,\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\left(x\ne-1;x\ne3\right)\\ \Leftrightarrow\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{x}{2\left(3-x\right)}\\ \Leftrightarrow\dfrac{x\left(x-3\right)-2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\\ \Leftrightarrow x^2-3x-4x=-x^2-x\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(c,\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Leftrightarrow-2x^2+x+1=2x^2-2x\\ \Leftrightarrow4x^2-3x-1=0\\ \Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(d,\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}=\dfrac{5-x}{2x^2+10x}\left(x\ne5;x\ne-5\right)\\ \Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\\ \Leftrightarrow\dfrac{x^2+25x-2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x+5\right)\left(x-5\right)}\\ \Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=-\left(x^2-10x+25\right)\\ \Leftrightarrow x^2+25x-2x^2-20x-50=-x^2+10x-25\\ \Leftrightarrow-5x=25\\ \Leftrightarrow x=-5\)
Tick nha
giúp mình nhe :>
A)3x^2-x(3x-5)=9
B)5x^2+9x-2=0
C)x/x+5-x-2/x=2x-1/x^2+5x
D)4(5-3x)=5x-5
E)2x^2-11x+14=0
F)3/2x+3-5/x(2x+3)=4/x
A) 3x² - x(3x - 5) = 9
3x² - 3x² + 5x = 9
5x = 9
x = 9/5
--------------------
B) 5x² + 9x - 2 = 0
5x² + 10x - x - 2 = 0
(5x² + 10x) - (x + 2) = 0
5x(x + 2) - (x + 2) = 0
(x + 2)(5x - 1) = 0
x + 2 = 0 hoặc 5x - 1 = 0
*) x + 2 = 0
x = -2
*) 5x - 1 = 0
5x = 1
x = 1/5
Vậy x = -2; x = 1/5
---------------------
D) 4(5 - 3x) = 5x - 5
20 - 12x = 5x - 5
-12x - 5x = -5 - 20
-17x = -25
x = 25/17
--------------------
E) 2x² - 11x + 14 = 0
2x² - 4x - 7x + 14 = 0
(2x² - 4x) - (7x - 14) = 0
2x(x - 2) - 7(x - 2) = 0
(x - 2)(2x - 7) = 0
x - 2 = 0 hoặc 2x - 7 = 0
*) x - 2 = 0
x = 2
*) 2x - 7 = 0
2x = 7
x = 7/2
Vậy x = 2; x = 7/2
Câu C và F ghi đề bằng công thức đúng lại em
c: =>x^2-(x-2)(x+5)=2x-1
=>x^2-x^2-5x+2x+10=2x-1
=>3x+10=2x-1
=>x=-11
f: =>3x-5=4(2x+3)
=>8x+12=3x-5
=>5x=-17
=>x=-17/5
Giải các bất phương trình sau và biểu diễn tập nghiệm trên trục số:
a.3x-5 >15-x b.3(x-2).(x+2)<3x^2+x
c.(2x+1)^2+(1-x).3x<hoặc=(x+2)^2
d.5x-20/3 - 2x^2+x/2 > x.(1-3x)/3 -5x/4
e.4-2x <hoặc= 3x-6
f.(x+4).(5x-1)>5x^2+16x+2
g)x.(2x-1)-8<5-2x(1-x)
h)3x-1/4 - 3.(x-2)/8 - 1>5-3x/2
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
1) (x+6)(3x-1)+x+6=0
2) (x+4)(5x+9)-x-4=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
4)2x (2x-3)=(3-2x)(2-5x)
5)(2x-7)^2-6(2x-7)(x-3)=0
6)(x-2)(x+1)=x^2-4
7) x^2-5x+6=0
8)2x^3+6x^2=x^2+3x
9)(2x+5)^2=(x+2)^2
1) (x+6)(3x-1)+x+6=0
⇔(x+6)(3x-1)+(x+6)=0
⇔(x+6)(3x-1+1)=0
⇔3x(x+6)=0
2) (x+4)(5x+9)-x-4=0
⇔(x+4)(5x+9)-(x+4)=0
⇔(x+4)(5x+9-1)=0
⇔(x+4)(5x+8)=0
3)(1-x)(5x+3)÷(3x-7)(x-1)
=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)