BM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
16 tháng 12 2022
a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)
\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)
=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0
=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0
=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0
=>(2x^2+120+35x)(2x^2+31x+120)=0
=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)
b: Đặt x^2-3x=a
Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)
\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)
=>(3a+10)(a+5)=6(a^2+7a+12)
=>6a^2+42a+72=3a^2+15a+10a+50
=>3a^2+17a+22=0
=>x=-2 hoặc x=-11/3
28 tháng 3 2020
ĐKXĐ bạn tự tìm nha : )
k, Ta có : \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}\)
\(=\frac{3x\left(1-2x\right)\left(1+2x\right)}{2x\left(x+4\right)\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)
j, Ta có : \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}:\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}=\frac{x+y}{y-x}.\frac{3\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\)
\(=\frac{3\left(x-y\right)\left(x+y\right)}{x\left(y-x\right)}=\frac{3\left(x-y\right)\left(x+y\right)}{-x\left(x-y\right)}=\frac{-3\left(x+y\right)}{x}\)
i, Ta có : \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right)}{-\left(a-b\right)}:\frac{a+b}{2\left(a^2-b^2\right)}=\frac{a\left(a+b\right)}{-\left(a-b\right)}.\frac{2\left(a-b\right)\left(a+b\right)}{a+b}\)
\(=\frac{2a\left(a+b\right)\left(a-b\right)}{-\left(a-b\right)}=-2a\left(a+b\right)\)
h, = k,
f, Ta có : \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{-3}{x-6}=\frac{-3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(x-6\right)}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
2 tháng 6 2022
Câu 1:
\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)
AH
Akai Haruma
Giáo viên
12 tháng 8 2020
f)
$\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{(2-3x)^3}$
$=\frac{2x-3x^2}{x^2-1}.\frac{x^4-1}{(2-3x)^3}=\frac{x(2-3x)(x^2-1)(x^2+1)}{(x^2-1)(2-3x)^3}$
$=\frac{x(x^2+1)}{(2-3x)^2}$
g)
$\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}=\frac{5xy}{2x-3}.\frac{12-8x}{15xy^3}$
$=\frac{5xy}{2x-3}.\frac{-4(2x-3)}{15xy^3}=\frac{-4}{3y^2}$
h)
$\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}=\frac{x(x+2)}{3(x-1)^2}:\frac{2(x+2)}{5(x-1)}$
$=\frac{x(x+2)}{3(x-1)^2}.\frac{5(x-1)}{2(x+2)}$
$=\frac{5x}{6(x-1)}$
AH
Akai Haruma
Giáo viên
12 tháng 8 2020
d)
$\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}=\frac{x+8}{(x-4)(x+4)}-\frac{2}{x(x+4)}$
$=\frac{x(x+8)}{x(x-4)(x+4)}-\frac{2(x-4)}{x(x+4)(x-4)}$
$=\frac{x^2+8x-2(x-4)}{x(x+4)(x-4)}=\frac{x^2+6x+8}{x(x+4)(x-4)}$
$=\frac{(x+2)(x+4)}{x(x+4)(x-4)}=\frac{x+2}{x(x-4)}$
e)
$\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{(x-7)(x+7)}{2x+1}.\frac{-3}{x-7}$
$=\frac{-3(x+7)}{2x+1}$
10 tháng 7 2022
a: \(\Leftrightarrow2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
=>5x=22
hay x=22/5
b: \(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
hay \(x\in\left\{3;-\dfrac{11}{10}\right\}\)
c: \(\Leftrightarrow x^3+2x^2-5x-10+5x=2x^2+17\)
\(\Leftrightarrow x^3+2x^2-10-2x^2-17=0\)
=>x3=27
=>x=3
d: \(\Leftrightarrow x^3+1-x^3+3x=15\)
=>3x=14
hay x=14/3
CM
4 tháng 12 2018
a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)
\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)
\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)
\(=-2x^2+2x+6\)
\(=-2\left(x^2-x-3\right)\)
b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)
\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)
\(=x^4+4x^2+4-x^4+16\)
\(=4x^2+20\)
\(=4\left(x^2+5\right)\)
c) \(\left(2x-y\right)^2-2\left(x+3y\right)^2-\left(1+3x\right)\left(3x-1\right)\)
\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)
\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)
\(=-7x^2-20xy-17y^2+1\)
d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^4+3x^2\)
\(=-3x^2\left(x^2-1\right)\)
\(=-3x^2\left(x-1\right)\left(x+1\right)\)
e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)
\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)
\(=\left(2x-1-2x-1\right)^2\)
\(=\left(-2\right)^2=4\)
g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y+z\right)^2\)
\(=\left(x+2z\right)^2\)
h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)
\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)
\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)
\(=5x^2+2x^2+3x-1-3x-3\)
\(=7x^2-4\)
KB
8 tháng 9 2018
a ) \(A=3x^2+5x-2\)
\(=3\left(x^2+\dfrac{5}{3}x-\dfrac{2}{3}\right)\)
\(=3\left(x^2+2x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{49}{36}\right)\)
\(=3\left[\left(x+\dfrac{5}{6}\right)^2-\dfrac{49}{36}\right]\)
\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{49}{12}\ge-\dfrac{49}{12}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{5}{6}=0\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy Min A là : \(-\dfrac{49}{12}\Leftrightarrow x=-\dfrac{5}{6}\)
b ) \(B=3x^2-4x+1\)
\(=3\left(x^2-\dfrac{4}{3}x+\dfrac{1}{3}\right)\)
\(=3\left(x^2-2x.\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{1}{9}\right)\)
\(=3\left[\left(x-\dfrac{2}{3}\right)^2-\dfrac{1}{9}\right]\)
\(=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{1}{3}\ge-\dfrac{1}{3}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{2}{3}=0\Leftrightarrow x=\dfrac{2}{3}\)
Vậy Min B là : \(-\dfrac{1}{3}\Leftrightarrow x=\dfrac{2}{3}\)
c ) \(C=-x^2-3x-2\)
\(=-\left(x^2+3x+2\right)\)
\(=-\left(x^2+2x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}\right)\)
\(=-\left[\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\right]\)
\(=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy Max C là : \(\dfrac{1}{4}\Leftrightarrow x=-\dfrac{3}{2}\)
d ) \(D=-4-3x^2+2x\)
\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{4}{3}\right)\)
\(=-3\left(x^2-2x.\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)
\(=-3\left[\left(x-\dfrac{1}{3}\right)^2+\dfrac{11}{9}\right]\)
\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{11}{3}\le-\dfrac{11}{3}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy Max D là : \(-\dfrac{11}{3}\Leftrightarrow x=\dfrac{1}{3}\)
:D
TN
8 tháng 9 2018
A=3x2+5x-2=2x2+4x+2+x2 -4=2(x+1)2 +x2-4 >=-4
Vậy A min=-4
B=3x2-4x+1=2x2-4x+2+x2-1=2(x-1)2+x2-1>=-1
Vậy B min=-1
C=-x2-3x -2=-(x2+3x+2)=-(x2+2x.3/2+9/4-1/4)=-(x+3/2)2+1/4
Ta có -(x+3/2)2<=0
=>-(x+3/2)2+1/4<=1/4
=> C max=1/4
D=-4-3x2+2x=-3x2+2x-4=-3(x2-2x/3+4/3)
=-3(x2-2x.1/3+1/9+11/9)=-3(x-1/3)2-11/3
Ta có -3(x-1/3)2<=0
=>-3(x-1/3)2-11/3<=-11/3
Vậy D max=-11/3
Đề là gì vậy bạn? Phân tích đa thức thành nhân tử à?