a, /2x^2-5x+4/=2x-1
b, √2+3x-x^2=3x-4
c,, /3x^2-4x+1/=/3x-1/
d, 2-√3x^2-9x+1=x
Mong mọi người giúp đỡ và các thầy cô giáo
a, √21-x^2-4x=x+3
b, /x^2+5x+4/-4=x
c, /2x^2-5x+4/=2x-1
d, √2+3x-x^2=3x-4
e, /3x^2-4x+1/=/3x-1/
f, 2-√3x^2-9x+1=x
Mong mọi người giúp đỡ và các thầy cô giáo
a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\x^2+6x+9=21-x^2-4x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\2x^2+10x-12=0\end{matrix}\right.\Leftrightarrow x=1\)
b: \(\left|x^2+5x+4\right|-4=x\)
=>|x^2+5x+4|=x+4
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-4\\\left(x^2+5x+4-x-4\right)\left(x^2+5x+4+x+4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-4\\\left(x^2+4x\right)\left(x^2+6x+8\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;-2;-4\right\}\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x^2-5x+4-2x+1\right)\left(2x^2-5x+4+2x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x^2-7x+5\right)\left(2x^2-3x+3\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{5}{2};1\right\}\)
1.Giải phương trình:
a) 4x-8/2x^2+1 = 0
b)x^2-x-6/x-3 = 0
c)x+5/3x-6 - 1/2 = 2x-3/2x-4
d)12/1-9x^2 = 1-3x/1+3x - 1+3x/1-3x
2.Giải các phương trình:
a)5 + 96/x^2-16 = 2x-1/x+4 - 3x-1/4-x
b)3x+2/3x-2 - 6/2+3x = 9x^2/9x^2-4
c)x+1/x^2+x+1 - x-1/x^2-x+1 = 3/x(x^4+x^2+1)
(mong mn giúp mk, mk đang thật sự gấp, cảm ơn mọi người rất nhiều)
\(a.\frac{4x-8}{2x^2+1}=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(b.\frac{x^2-x-6}{x-3}=0\left(x\ne3\right)\\\Leftrightarrow x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\\Leftrightarrow \left(x-3\right)\left(x+2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình trên là \(-2\)
\(c.\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\left(x\ne2\right)\\ \Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\\\Leftrightarrow \frac{2\left(x+5\right)}{6\left(x-2\right)}-\frac{3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\\\Leftrightarrow 2\left(x+5\right)-3\left(x-2\right)=3\left(2x-3\right)\\\Leftrightarrow 2x+10-3x+6=6x-9\\\Leftrightarrow 2x-3x-6x=-10-6-9\\\Leftrightarrow -7x=-25\\\Leftrightarrow x=\frac{25}{7}\left(tm\right)\)
Vậy nghiệm của phương trình trên là \(\frac{25}{7}\)
Bài 2: Tìm x, biết:
a/ 12x(x – 5) – 3x(4x - 10) = 120
b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)
c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`
Bài 1: Thực hiện phép tính
a) (3x-1)(9x2+3x+1)-4x(x-5)
b) (7x+2)(3-4x)-(x+3)(x2-3x+9)
c) (4x+3)(4x-3)-(2-x)(4+2x+x2)
d) (3x-8)(-5x+6)-(4x+1)(3x-2)
e) (3x-6)4x-2x(3x+5)-4x2
f) (5x-6)(6x-5)-x(3x+10)
Bài 2 : Tính
a) x(x+3)-x2=6
b) 2x(x-5)+x(-2x-1)=6
c) x (x+5)-(x+1)(x-2)=7
d)(3x+4)(6x-3)-(2x+1)(9x-2)=10
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
Giải Phương Trình
a, (2x+3)^2 - 3(x-4) (x+4) = (x-2)^2 +1
b, (3x-2) (9x^2 + 6x +4) - (3x-1) ( 9x^2 - 3x +1) = x+4
c, x(x-1) - (x-3) (x+4) = 5x
d, (2x+1) (2x-1) = 4x(x-7) - 3x
B) (2x+3)2-(5x-4) (5x+4)=(x+5)2-(3x-1) (7x+2)-(x2-x+1)
C) (1-3x)2-(x-2) (9x+1)=(3x-4) (3x+4)-9(x+3)2
D) (3x+4) (3x-4) - (2x+5)2=(x-5)2+(2x+1)2-(x2-2x)+(x-1)2 cần gắp
tìm x biết
A) (4x-1)2-(3x+2) (3x-2)=(7x-1) (x+2) + (2x+1)2-(4x1+7)
B) (2x+3)2-(5x-4) (5x+4)=(x+5)2-(3x-1) (7x+2)-(x2-x+1)
C) (1-3x)2-(x-2) (9x+1)=(3x-4) (3x+4)-9(x+3)2
D) (3x+4) (3x-4) - (2x+5)2=(x-5)2+(2x+1)2-(x2-2x)+(x-1)2 cần gắp
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
Tìm x,biết
b) |3/4x-5|-2/3=|-1/4|
a) (3x-1) (-1/4 - 5x) (-2/7x + 3)=0
c)|-2/5 - 3x| - |-7/9 + 2x|=0
d) |-3/7x - 1| + |5+2/3x|=0
e)|3x-1|+|9x^2 - 1|=0
f) x+2/2018 + x+4/2016 = x+6/2014 - 1
Mọi người giải giúp e với ạ chiều e phải nộp r
ngu thế à bạn
Tìm x, biết:
a)2x*(6x-5)-4x*(3x+7)=7
b)-5x(2x+1)+3x(3x+2)+x(x-1/2)=0
c)3x*(6x-5)-2x(9x+7)=15
d)1/2x*(2x+4)-(x+3)=5
e)(-3x+2)*5x-5x*(2x+1)-5x=4
Mấy bạn giúp mk ik mk đang cần gấp!
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)