Tính:
2x+1/2^2-x + -32x^2/(2x+1)(2x-1) + 1-2x/x(2x+1)
tìm x:
a)(2x-3)+(3x^2+1)-6x*(x^2-x+1)+3x^2-2x=10
b)(3x+1)*(x-2)-x*((3x-5)=-8-5x
c)(4x-3)*(16x^2+12+9)-32x^2*(2x-1)-32x^2+x=20
a: \(\left(2x-3\right)\left(3x^2+1\right)-6x\left(x^2-x+1\right)+3x^2-2x=10\)
\(\Leftrightarrow6x^3+2x-9x^2-3-6x^3+6x^2-6x+3x^2-2x=10\)
\(\Leftrightarrow-6x-3=10\)
=>-6x=13
hay x=-13/6
b: \(\Leftrightarrow3x^2-3x+x-2-3x^2+5x=-8-5x\)
=>3x-2=-5x-8
=>8x=-6
hay x=-3/4
c: \(\Leftrightarrow64x^3-27-64x^3+32x^2-32x^2+x=20\)
=>x-27=20
hay x=47
1) (1-x)(5x+3)=(3x-7)(x-1)
2) (x-2)(x+1)=x2-4
3) 2x3+3x2-32x=48
4) x2+2x-15=0
5) 2x(2x-3)=(3-2x)(2-5x)
6) x3-5x2+6x=0
7) (x2-5)(x+3)=0
8) (x+7)(3x-1)=49-x2
\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)
\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)
\(< =>\left(1-x\right)\left(8x-4\right)=0\)
\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
\(\left(x-2\right)\left(x+1\right)=x^2-4\)
\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)
\(< =>-1\left(x-2\right)=0\)
\(< =>2-x=0< =>x=2\)
\(2x^3+3x^2-32x=48\)
\(< =>x^2\left(2x+3\right)-16\left(2x+3\right)=0\)
\(< =>\left(x^2-16\right)\left(2x+3\right)=0\)
\(< =>\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)
\(< =>\hept{\begin{cases}x=4\\x=-4\\x=-\frac{3}{2}\end{cases}}\)
Làm tính cộng các phân thức
a) 11x+13/3x-3 + -15x-17/4x-4
b) 2x+1/2x2-x + -32x2/4x2-1 + 1-2/2x2+x
Làm tính cộng các phân thức :
a) \(\dfrac{11x+13}{3x-3}+\dfrac{15x+17}{4-4x}\)
b) \(\dfrac{2x+1}{2x^2-x}+\dfrac{32x^2}{1-4x^2}+\dfrac{1-2x}{2x^2+x}\)
c) \(\dfrac{1}{x^2+x+1}+\dfrac{1}{x^2-x}+\dfrac{2x}{1-x^3}\)
d) \(\dfrac{x^4}{1-x}+x^3+x^2+x+1\)
Giải pt
1, 9x^2-1 =(3x+1)(4x+1)
2, 2x^3+3x^2-32x=48
3, (2x+5)^2-(x+2)^2=0
4, 2x^3+6x^2=x^2+3x
a) \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\)\(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
Vậy...
Bài cộng phân thức
1. a) \(\frac{11x+13}{3x-3}+\frac{15x+17}{4-4x}\)
b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}\)
Bài cộng phân thức
1. a) \(\frac{11x+13}{3x-3}+\frac{15x+17}{4-4x}\)
b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}\)
really your name is crazy
and do you crazy
Giải phương trình:
a) √x+2+3√2x−5+√x−2−√2x−5=2√2x+2+32x−5+x−2−2x−5=22
b) √x+1+2(x+1)=x−1+√1−x+3√1−x2x+1+2(x+1)=x−1+1−x+31−x2
c) (√1+x−1)(√1−x+1)=2x(1+x−1)(1−x+1)=2x
d) 2(x2+2x+3)=5√(x+1)32(x2+2x+3)=5(x+1)3
(Các bạn giúp mình sớm nhé! Thank!)
Bài 1. Tính:
32x^m. 1/2x
(a+5).4
(3a-5b).2a
(a^m+2a^3). a^n
x(2x+1)
-6x+3.(7+2x)
Bài 2. Tìm x: 3x+2(5-x)=0
Bài 3. Rút gọn biểu thức
6(3p+4q)-8(5p-q)+(p-q).
Bài 3. tính giá trị biểu thức sau khi rút gọn:
5x(4x^2-2x+1)- 2x(10x^2-5x-2) với x= -15.
Bài 2:
3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = 0 - 10
<=> x = -10
=> x = -10
Bài 3:
6(3q + 4q) - 8(5p - q) + (p - q)
= 6.3p + 6.4q - 8.5p - (-8).q + p - q
= 18p + 24q - 40p + 8q + p - q
= (18p - 40p + p) + (24q + 8q - q)
= -21p + 31q
Bài 4:
5x(4x2 - 2x + 1) - 2x(10x2 - 5x - 2)
= 5x.4x2 + 5x.(-2x) + 5x.1 - 2x.10x2 + (-2x).(-5x) + (-2x).(-2)
= 20x3 - 10x2 + 5x - 20x3 + 10x2 + 4x
= (20x2 - 20x2) + (-10x2 + 10x2) + (5x + 4x)
= 0 + 0 + 9x
= 9x (1)
Thay x = -15 vào (1), ta có:
9.(-15) = -135
Vậy: Giá trị biểu thức sau khi rút gọn với x = -15 là: -135