|2x−1|+|x^2 −3x+2| ≥ x−2
Rút gọn :
1. (2x-5)(3x+1)-(x-3)^2+(2x+5)^2-(3x+1)^3
2. (2x-1)(2x+1)-3x-2)(2x+3)-(x-1)^3+(2x+3)^3
3. (x-2)(x^2+2x+4)-(3x-2)^3+(3x-4)^2
4. (7x-1)(8x+2)-(2x-7)^2-(x-4)^3-(3x+1)^3
5. (5x-1)(5x+1)-(x+3)(x^2-3x+9)-(2x+4)^2-(3x-4)^2+(2x-5)^3
6. (4x-1)(x+2)-(2x+5)^2-(3x-7)^2+(2x+3)^3=(3x-1)^3
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
1) |2x - 1| = 5
2) |2x - 1| = |x + 5|
3) |3x + 1| = x - 2
4) |3 - 2x| = x + 2
5) |2x - 1| = 5 - x
6) |- 3x| = x - 2
7) |2 - 3x| = 2x + 1
8) |2x - 1| + |4x ^ 2 - 1| = 0
9) (2x + 5)/(x + 3) + 1 = 4/(x ^ 2 + 2x - 3) - (3x - 1)/(1 - x)
10) (x - 1)/(x + 3) - x/(x - 3) = (7x - 3)/(9 - x ^ 2)
11) 5 + 96/(x ^ 2 - 16) = (2x - 1)/(x + 4) + (3x - 1)/(x - 4)
12) (2x)/(2x - 1) + x/(2x + 1) = 1 + 4/((2x - 1)(2x + 1))
13) (x + 2)/(x - 2) - 1/x = 2/(x ^ 2 - 2x)
14) x/(2x - 6) + x/(2x + 2) = (2x + 4)/(x ^ 2 - 2x - 3)
Rút gọn biểu thức:
a) (x 2 – 2x + 2)(x 2 – 2)(x 2 + 2x + 2)(x 2 + 2)
b) (x + 1)2 – (x – 1)2 + 3x 2 – 3x(x + 1)(x – 1)
c) (2x + 1)2 + 2(4x 2 – 1) + (2x – 1)2
d) (m + n)2 – (m – n)2 + (m – n)(m + n)
e) (3x + 1)2 – 2(3x + 1)(3x + 5) + (3x + 5)2
a: Ta có: \(\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
\(=\left(x^4-4\right)\left[\left(x^2+2\right)^2-4x^2\right]\)
\(=\left(x^4-4\right)\left(x^4+4x^2+4-4x^2\right)\)
\(=\left(x^4-4\right)\cdot\left(x^4+4\right)\)
\(=x^8-16\)
b: Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2+3x^2-3x\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-x^2+2x-1+3x^2-3x\left(x^2-1\right)\)
\(=3x^2+4x-3x^3+3x\)
\(=-3x^3+3x^2+7x\)
a) x\(^2\)-3x+7=1+2x
b) x\(^2\)-3x-10=0
c) x\(^2\)-3x+4=2(x-1)
d) (x+1)(x-2)(x-5)=0
e) 2x\(^2\)+3x+1=0
f) 4x\(^2\)-3x=2x-1
a) Ta có: \(x^2-3x+7=1+2x\)
\(\Leftrightarrow x^2-3x+7-1-2x=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: S={3;2}
b) Ta có: \(x^2-3x-10=0\)
\(\Leftrightarrow x^2-5x+2x-10=0\)
\(\Leftrightarrow x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy: S={5;-2}
c) Ta có: \(x^2-3x+4=2\left(x-1\right)\)
\(\Leftrightarrow x^2-3x+4=2x-2\)
\(\Leftrightarrow x^2-3x+4-2x+2=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: S={3;2}
d) Ta có: \(\left(x+1\right)\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=5\end{matrix}\right.\)
Vậy: S={-1;2;5}
e) Ta có: \(2x^2+3x+1=0\)
\(\Leftrightarrow2x^2+2x+x+1=0\)
\(\Leftrightarrow2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{-1}{2}\right\}\)
f) Ta có: \(4x^2-3x=2x-1\)
\(\Leftrightarrow4x^2-3x-2x+1=0\)
\(\Leftrightarrow4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{1;\dfrac{1}{4}\right\}\)
Bài 1:
a) x (\(x^2\) + 2) + 2x\((1-\dfrac{1}{2}x^2)=4\)
b) (2x)\(^2\) (x – 1) + x(\(x^2\) + 4x) = 40
c) 3x(x – 2) – 3(\(x^2\) – 3) = 8
d) 2\(x^2\)(4\(x^3\) + 2x) + (\(x^2\) – 2)(- 2x)\(^3\) = 20
Bài 2:
P = 3x(\(\dfrac{2}{3}\)\(x^2\) − \(3x^4)\) + (3x)\(^2\) (\(x^3\) – 1) + (- 2x + 9)\(x^2\) - 12
Bài 2:
Ta có: \(P=3x\left(\dfrac{2}{3}x^2-3x^4\right)+9x^2\left(x^3-1\right)+x^2\left(-2x+9\right)-12\)
\(=2x^3-9x^5+9x^5-9x^2-2x^3+9x^2-12\)
=-12
Bài 1:
a: Ta có: \(x\left(x^2+2\right)+2x\left(1-\dfrac{1}{2}x^2\right)=4\)
\(\Leftrightarrow x^3+2x+2x-x^3=4\)
hay x=1
b: Ta có: \(4x^2\left(x-1\right)+x\left(x^2+4x\right)=40\)
\(\Leftrightarrow4x^3-4x^2+x^3+4x^2=40\)
\(\Leftrightarrow5x^3=40\)
hay x=2
c: Ta có: \(3x\left(x-2\right)-3\left(x^2-3\right)=8\)
\(\Leftrightarrow3x^2-6x-3x^2+9=8\)
\(\Leftrightarrow-6x=-1\)
hay \(x=\dfrac{1}{6}\)
tìm x
a,(3x-4)(3x+4)-(3x+1)2=0
b,(2x-5)2-(2x+1)(2x-1)=0
c,(3x-1)2-2(3x-1)(-3x)+(x-3)2=25
d,(x-1)(x2+x+1)=7
e,(x+2)(x2-2x+4)=-19
giút gọn
a,(x-1)(x2+x+1)-(x+2)(x2-2x+4)
b,(x-2)(x2+2x+4)-(x+2)(x-2)-x(x2-2)
c,(2x-1)(4x2+2x+1)+(2x+1)(4x2-2x+1)
trời đất, học hằng đẳng thức chưa, chưa hc thì thôi, học rồi thì áp dụng vs bài này như ăn cháo thôi chứ có j đâu phải hỏi
A= ( x+2)(x2-2x+4) - (x3-2)
B= (x+2)(x-2)(x2+2x+4)(x2+2x+4)(x2- 2x+4)
C= (x2+3x+1)2+(3x-1)2-2(x2+3x+1)(3x-1)
D= (3x3+3x+1)(3x2-3x+1)-(3x3+1)2
E= (2x2+2x+1)(2x2-2x+1)-(2x2+1)2
Giúp em với ạ.
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(\Rightarrow A=\left(x^3+8\right)-\left(x^3-2\right)\)
\(\Rightarrow A=x^3+8-x^3+2\)
\(\Rightarrow A=\left(x^3-x^3\right)+\left(8+2\right)\)
\(\Rightarrow A=10\)
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(=x^3+8-x^3+2\)
\(=10\)
\(B=\left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x^3+8\right)\left(x^3-8\right)\)
\(=x^6-64\)
\(C=\left(x^2+3x+1\right)^2+\left(3x-1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)\)
\(=\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x+1-3x+1\right)^2\)
\(=\left(x^2+2\right)^2\)
\(D=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1+3x\right)\left(3x^3+1-3x\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)
\(=-9x^2\)
\(E=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1\right)^2-4x^2-\left(2x^2+1\right)^2\)
\(=-4x^2\)
giải pt
a 2(x+3)(x-4)=(2x-1)(x+2)-27
b (3x+2)(x-1)-3(x+1)(x-2)=4
c (x+2)(x^2 -2x+4)-x(x-3)(x+3)=26
d (3x+2)(3x-2)-(3x-4)^2=28
e 5(x+3)^2-5(x-4)(x+8)=3x
f 2x(x+2)^2-8x^2=2(x-2)(x^2+2x+4)
g (2x-1)(4x^2+2x+1)-4x(2x^2-3)=23
h x(x-2)(x+2)-(x-3)(x^2+3x+9)+1=0
i x(x^2+x+1)-(x-1)(x+1)x=x^2+2
a, \(2\left(x+3\right)\left(x-4\right)=\left(2x-1\right)\left(x+2\right)-27\)
\(\Leftrightarrow2\left(x^2-4x+3x-12\right)=2x^2+4x-x-2-27\)
\(\Leftrightarrow2x^2-2x-24=2x^2+3x-29\Leftrightarrow-5x+5=0\Leftrightarrow x=1\)
b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3-8-x\left(x^2-9\right)=26\Leftrightarrow-8+9x=26\)
\(\Leftrightarrow9x=18\Leftrightarrow x=2\)
Bài 3. Rút gọn các đa thức sau
a/ (2x-3)(4x^2+6x+9)- (2x+1)(4x^2 - 2x +1)
b/ (x+ 2)(x^2- 2x+4) – (x^3- 2)
c/ (3x+ 5)(9x^2 - 15x +25)- 3x(3x-1)(3x+1)
d/ x^6 - (x^2 + x +1)(x^2 - 1)(x^2 - x+ 1)
a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28
b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10
c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x
d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1
Bài 1 : giải phương trình
a) (x-2)(x+2)-(2x+1)2=x(2-3x)
b) 2x(x+2)2-8x2=2(x-2)(x2+2x+4)
c) (x-2)3+(3x-1)(3x+1)=(x+1)3
d) 5(2x-3)-4(5x-7)=19-2(x+1)2
a: \(\Leftrightarrow x^2-4-4x^2-4x-1-2x+3x^2=0\)
=>-6x-5=0
=>-6x=5
hay x=-5/6
b: \(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)
=>8x+16=0
hay x=-2
c: \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)
=>9x-10=0
hay x=10/9
d: \(\Leftrightarrow10x-15-20x+28=19-2x^2-4x-2\)
\(\Leftrightarrow-10x+13+2x^2+4x-17=0\)
\(\Leftrightarrow2x^2-6x-4=0\)
\(\Leftrightarrow x^2-3x-2=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-2\right)=9+8=17>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{17}}{2}\\x_2=\dfrac{3+\sqrt{17}}{2}\end{matrix}\right.\)