Rút gọn
a) x(x+1)(x-1)-(x-1)(x2+x+1)
b) (9x-1)2+(1-5x)2+2(9x-1)(1-5x)
c) (a+b+c)(a2+b2+c2-ab-bc-ca)
bài 1: rút gọn bthuc
a.\(\dfrac{a+\sqrt{a}}{\sqrt{a}}\) b.\(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}\)
b2: rút gọn
a.\(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}\) b.4-x-\(\sqrt{4-4x+x^2}\) c.\(\sqrt{4x^2-4x\text{x^2 +2*x-3 >0}}-\sqrt{4x^2+4x+1}\)
Bài 1:
a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)
b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)
Bài 2:
a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)
b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)
Câu 1. Phân tích đa thức thành nhân tử
a) x 2 + 4xy + 3y2
b) x 3 – y 3 + z3 + 3xyz
c) x 4 + 2x2 – x + 2
Câu 2. Chứng minh rằng a = b = c nếu có một trong các điều kiện sau:
a) a 2 + b2 + c2 = ab + bc + ca
b) (a + b + c)2 = 3(a2 + b2 + c2 )
c) (a + b + c)2 = 3(ab + bc + ca)
Câu 3. Chứng minh rằng với số tự nhiên n thì A = n(n+1)(n+2)(n+3) + 1 là số chính phương.
Câu 4. Tìm x thỏa mãn a) (x – 1)3 + (x – 3)3 = (2x – 4)3 b) (2x – 1)3 + (x + 3)3 = (3x + 2)3 c) (2x + 1)3 + (3x + 3)3 + (-5x - 4)3 = 0
b2 tìm x
a)x^2-4x-5=0
b)5x^2-9x-2=0
c)(x^2+1)-5(x^2+1)+6=0
d)(x^2+6x)-2(x+3)^2-17=0
Lời giải:
a. $x^2-4x-5=0$
$\Leftrightarrow (x+1)(x-5)=0$
$\Leftrightarrow x+1=0$ hoặc $x-5=0$
$\Leftrightarrow x=-1$ hoặc $x=5$
b.
$5x^2-9x-2=0$
$\Leftrightarrow (x-2)(5x+1)=0$
$\Leftrightarrow x-2=0$ hoặc $5x+1=0$
$\Leftrightarrow x=2$ hoặc $x=\frac{-1}{5}$
c.
$(x^2+1)-5(x^2+1)+6=0$
$\Leftrightarrow a^2-5a+6=0$ (đặt $x^2+1=a$)
$\Leftrightarrow (a-2)(a-3)=0$
$\Leftrightarrow a-2=0$ hoặc $a-3=0$
$\Leftrightarrow x^2-1=0$ hoặc $x^2-2=0$
$\Leftrightarrow (x-1)(x+1)=0$ hoặc $(x-\sqrt{2})(x+\sqrt{2})=0$
$\Leftrightarrow x\in\left\{\pm 1; \pm \sqrt{2}\right\}$
d.
$(x^2+6x)-2(x+3)^2-17=0$
$\Leftrightarrow (x^2+6x+9)-2(x+3)^2-26=0$
$\Leftrightarrow (x+3)^2-2(x+3)^2-26=0$
$\Leftrightarrow -(x+3)^2-26=0$
$\Leftrightarrow (x+3)^2=-26<0$ (vô lý)
Do đó không tồn tại $x$ thỏa mãn.
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
\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)
\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)
\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)
\(=-10x^3+19x^2+74x+1\)
\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)
\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)
\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)
\(=-5x^4-11x^3+24x^2+12x+7\)
\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)
\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)
\(=-2x^2-27x+57\)
\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)
\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)
\(=-x^3+4x^2+22x+5\)
\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)
\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)
\(=-9x^3-55x^2+4x+35\)
\(g,\left(x-1\right)^2-\left(x+2\right)^2\)
\(=x^2-2x+1-x^2-4x-4\)
\(=-6x-3\)
Rút gọn biểu thức
a. 2x+2y/a2+2ab+b2 . ax-ay+bx-by/2x2-2y2
b. a+b-c/a2+2ab+b2-c2 . a2+2ab+b2+ac+bc/a2-b2
c.x3+1/x2+2x+1 . x2-1/2x2-2x+2
d. x8-1/x+1 . 1/ (x2+1) (x4+1)
e. x-y/xy+y2 - 3x+y/x2-xy . y-x/x+y
a2 c2... là em viết số mũ đó ạ. anh chị giúp em giải mấy bài này nha
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)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
Bài 1: rút gọn
a, (2x-1)^2 +4(x-1) (x+3)-2(5-3x)^2
b, (2a^2+2a+1)(2a^2-2a+1)-(2a^2+1)^2
c, (9x-1)^2+(1-5x)^2+2(9x-1)(1-5x)
d, (x^2+5x-1)^2+2(5x-1)(x^2-5x+1)+(5x-1)^2
e, x(x-1)(x+1)-(x+1)(x^2-x+1)
f, x(x+4)(x-4)-(x^2+1)(x^2-1)
Bài 1:
a) Ta có: \(\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=4x^2-4x+1+4\left(x^2+2x-3\right)-2\left(25-30x+9x^2\right)\)
\(=4x^2-4x+1+4x^2+8x-12-50+60x-18x^2\)
\(=-10x^2+64x-61\)
b) Ta có: \(\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a^2+1\right)^2\)
\(=\left(2a^2+1\right)^2-\left(2a\right)^2-\left(2a^2+1\right)^2\)
\(=-4a^2\)
c) Ta có: \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
\(=\left(9x-1+1-5x\right)^2\)
\(=\left(4x\right)^2=16x^2\)
d)
Sửa đề: \(\left(x^2+5x-1\right)^2+2\left(5x-1\right)\left(x^2+5x-1\right)+\left(5x-1\right)^2\)
Ta có: \(\left(x^2+5x-1\right)^2+2\left(5x-1\right)\left(x^2+5x-1\right)+\left(5x-1\right)^2\)
\(=\left(x^2+5x-1+5x-1\right)^2\)
\(=\left(x^2+10x-2\right)^2\)
\(=x^4+100x^2+4+20x^3-40x-4x^2\)
\(=x^4+20x^3+96x^2-40x+4\)
e) Ta có: \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x\left(x^2-1\right)-\left(x^3+1\right)\)
\(=x^3-x-x^3-1\)
=-x-1
f) Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
bài 1 : Rút gọn biểu thức
a) 3x.(x-2)-5x(1-x)-8(x^2-3)
b) (7x-3) (2x+1) - (5x-2) (x+4)-9x^2 + 17x
\(a,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2+5x^2-8x^2\right)+\left(-6x-5x\right)+24\)
\(=0-11x+24\)
\(=-11x+24\)
\(b,\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x\)
\(=14x^2+7x-6x-3-5x^2-20x+2x+8-9x^2+17x\)
\(=\left(14x^2-5x^2-9x^2\right)+\left(7x-6x-20x+2x+17x\right)+\left(-3+8\right)\)
\(=0+0+5\)
\(=5\)
Rút gọn
a) (6x-2)2+(2+5x)2+4x(3x-12)x(2-5x)
b) (9x3+3x+1)x(9x3-3x+1)-(9x3+12)