Giải các phương trình sau:
a, x^2 + (x+2)(11x-7)=4
b, x^4+3x^2-4=0
c,(x-7)(x-5)(x-4)(x-2)=72
d, x(x+1)(x-1)(x+2)=4
e, 5x^3+6x^2+12x+8=0
Giải các phương trình sau:
a, x^2 + (x+2)(11x-7)=4
b, x^4+3x^2-4=0
c,(x-7)(x-5)(x-4)(x-2)=72
d, x(x+1)(x-1)(x+2)=4
e, 5x^3+6x^2+12x+8=0
a: \(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x^2-18=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)
=>x=-6 hoặc x=1
b: \(x^4+3x^2-4=0\)
\(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
106n+2 + 103n+1 +1
CMR : a) B chia hết cho 111 với n là số tự nhiên
b) B chia hết cho 91 với n lẻ
Giúp mình bài này với nhanh nhanh nhá cho a,b,c >0 và a+b+c=1 . Tìm giá GTNN của biểu thức P=1/a + 1/b + 1/c
Vì a;b;c > 0 nên \(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}>0\)
BĐT Cosi :
\(9a+\dfrac{1}{a}\ge2.\sqrt{9a.\dfrac{1}{a}}=2.3=6\\ 9b+\dfrac{1}{b}\ge6\\ 9c+\dfrac{1}{c}\ge6\\ \Rightarrow\left(9a+\dfrac{1}{a}\right)+\left(9b+\dfrac{1}{b}\right)+\left(9c+\dfrac{1}{c}\right)\ge18\\ \Rightarrow9\left(a+b+c\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge18\\ \Rightarrow9+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge18\\ \Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
Dấu "=" xảy ra khi a=b=c=1/3
Tìm m để phương trình (x-m)(x+2)-5mx+4=(x+m)(x-2)-3x có nghiệm gấp đôi nghiệm phương trình 2x(x-3)-6x=2(x-1)(x+5)
Giúp với!!!
b)Giải phương trình
\(\dfrac{x-a}{a+3}+\dfrac{x-3}{a-3}=\dfrac{6a}{9-a^2}\)
c)\(\dfrac{x-b-c}{a}+\dfrac{x-c-a}{b}+\dfrac{x-a-b}{c}=3vớia,b,c\ne0\)
d)\(\left(\dfrac{2011}{1\cdot11}+\dfrac{2011}{2\cdot12}+...+\dfrac{2011}{100\cdot110}\right)x=\dfrac{2012}{1\cdot101}+\dfrac{2012}{2\cdot102}+...+\dfrac{2012}{10\cdot110}\)
\(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\) hãy giải phương trình
\(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\Leftrightarrow\left(\dfrac{2-x}{2017}+1\right)=\left(\dfrac{1-x}{2018}+1\right)+\left(1-\dfrac{x}{2019}\right)\)
\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{2019-x}{2018}+\dfrac{2019-x}{2019}\)\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)
Ta đã có: \(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}< 0\)
Vậy ta dễ dàng suy ra được \(S=\left\{2019\right\}\)
-Cho t hỏi 1 chút nhá :D bài 6 trong sgk trang 9 ă
2) S= SABH+SBCKH+SCKD
= \(\dfrac{1}{2}.x.7+x^{2^{ }}+\dfrac{1}{2}+.x.4\)
=\(\dfrac{7x+2x^2+4x}{2}\) đến chỗ này tớ kh hiểu tại sao khi quy đồng lên x2 lại trở thành 2x2 chứ , bạn nào giải thích giùm mìh đi
Có MTC = 2 :
\(x^2=\dfrac{x^2}{1}=\dfrac{2x^2}{2}\)
Giai cac phuong trinh sau: a) x+1/65 + x+2/64= x+3/63 + x+4/62
b) x-12/77 + x-11/78 = x-74/15 + x-73/16
c) (1/1.2 + 1/2.3 +...+ 1/9.10) (x-1) + 1/10x = x-9/10
d) (1/1.51 + 1/2.52 +...+ 1/10.60)x = (1/1.4 + 1/2.12 +...+ 1/50.60)
a,\(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
\(\Rightarrow\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}\)
\(\Rightarrow\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}=\dfrac{x+3+63}{63}+\dfrac{x+4+62}{62}\)
\(\Rightarrow\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
\(\Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
\(\Rightarrow x+66=0\) ( vì \(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}>0\) )
\(\Rightarrow x=-66\)
\(b,\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Rightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1=\dfrac{x-74}{15}-1+\dfrac{x-73}{16}-1\)
\(\Rightarrow\dfrac{x-12-77}{77}+\dfrac{x-11-78}{78}=\dfrac{x-74-15}{15}+\dfrac{x-73-16}{16}\)
\(\Rightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)
\(\Rightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Rightarrow x-89=0\)
\(\Rightarrow x=89\)
b.
\(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\\ \Rightarrow\left(\dfrac{x-12}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)=\left(\dfrac{x-74}{15}-1\right)+\left(\dfrac{x-73}{16}-1\right)\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}=\dfrac{x-89}{15}+\dfrac{x-89}{16}\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\\ \\ \Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\\ \Leftrightarrow x-89=0\\ \Leftrightarrow x=89\)
a)giải phương trình
(8x+4x^2-1)(x^2+ 2x+1)=4(x^2+ x+1)
b)cho 2 số dương a,b thỏa mãn a+b#0.cmr a^2+b^2+(ab+1/a+b)»2
Giải phương trình:
a) \(\dfrac{2x^2-x-1}{2}-3x^2+x+4=\left(5-x\right)\left(2x+4\right)\)
b) \(\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\dfrac{\left(x-1\right)\left(2x+4\right)}{2}+1\)
c) 8x3 - 1 = 8x2 + 4x + 2
d) (x2 + x + 1)(x2 - x + 1) = x6 - 1
e) (x3 + 2x)(x2 + 4) = (x4 + 6x2 + 8)(3 - 2x)
f) \(\dfrac{x+97}{125}+\dfrac{x-63}{35}=\dfrac{x-7}{21}+\dfrac{x-27}{49}\)
Giúp mình với!!! Mình cần gấp!!! Câu nào cũng được nhe!!!
c) \(8x^3-1=8x^2+4x+2\)
<=> \(\left(2x-3\right)\left(4x^2+2x+1\right)=0\)
<=> \(2x-3=0\) hoặc \(4x^2+2x+1=0\)
Th1: x=\(\dfrac{3}{2}\)
Th2: Vô nghiệm
Vậy x=\(\dfrac{3}{2}\)
\(\text{a) }\dfrac{2x^2-x-1}{2}-3x^2+x+4=\left(5-x\right)\left(2x+4\right)\\ \Leftrightarrow\left(\dfrac{2x^2-x-1}{2}-3x^2+x+4\right)2=\left(5-x\right)\left(2x+4\right)2\\ \Leftrightarrow2x^2-x-1-6x^2+2x+8=\left(5-x\right)\left(4x+8\right)\\ \Leftrightarrow-4x^2+x+7=20x+40-4x^2-8x\\ \Leftrightarrow-4x^2+x+4x^2-12x=40-7\\ \Leftrightarrow-11x=33\\ \Leftrightarrow x=-3\\ \text{Vậy }S=\left\{-3\right\}\)
\(\text{b) }\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\dfrac{\left(x-1\right)\left(2x+4\right)}{2}+1\\ \Leftrightarrow\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\left(x-1\right)\left(x+2\right)+1\\ \Leftrightarrow\left(\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1\right)4=\left(x^2-x+2x-2+1\right)4\\ \Leftrightarrow\left(2x-5\right)\left(3x+7\right)+8x-4=\left(x^2+x-1\right)4\\ \Leftrightarrow6x^2-15x+14x-35+8x-4=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-39=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-4x^2-4x-39+4=0\\ \Leftrightarrow2x^2+3x-35=0\\ \Leftrightarrow2x^2+10x-7x-35=0\\ \Leftrightarrow\left(2x^2+10x\right)-\left(7x+35\right)=0\\ \Leftrightarrow2x\left(x+5\right)-7\left(x+5\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-5\end{matrix}\right.\\ \\ \text{Vậy }S=\left\{\dfrac{7}{2};-5\right\}\)
\(\text{c) }8x^3-1=8x^2+4x+2\\ \Leftrightarrow\left(2x-1\right)\left(4x^2+2x+1\right)=2\left(4x^2+2x+1\right)\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }S=\left\{\dfrac{3}{2}\right\}\)
\(\text{d) }\left(x^2+x+1\right)\left(x^2-x+1\right)=x^6-1\\ \Leftrightarrow\left(x^3+1\right)\left(x^3-1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x^2-x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x-1\right)=1\\ \Leftrightarrow x^2-1=1\\ \Leftrightarrow x^2=2\\ \Leftrightarrow x=\sqrt{2}\\ \text{Vậy }S=\left\{\sqrt{2}\right\}\)
\(\text{e) }\left(x^3+2x\right)\left(x^2+4\right)=\left(x^2+6x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+2x^2+4x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[\left(x^2+2x^2\right)+\left(4x^2+8\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[x^2\left(x^2+2\right)+4\left(x^2+2\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+4\right)\left(x^2+2\right)\left(3-2x\right)\\ \Leftrightarrow x=3-2x\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\\ \text{Vậy }S=\left\{1\right\}\)
f) Kiểm tra lại hạng tử thứ 2 ở vế phải.
Giải phương trình:
a) \(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
b) \(\dfrac{x-291}{1700}+\dfrac{x-293}{1698}+\dfrac{x-295}{1696}+\dfrac{x-297}{1694}=4\)
Giúp mình với!!! Mình cần gấp!!!
https://giaibaitapvenha.blogspot.com/2017/12/toan-lop-8-ai-so-giai-phuong-trinh.html
Nếu lời giải giúp ích được cho bạn, truy cập
https://giaibaitapvenha.blogspot.com/2017/12/en-voi-do-homework-for-you-e-trai.html
a,\(\dfrac{x-3}{113}+\dfrac{x-5}{115}=\dfrac{x-7}{117}+\dfrac{x-9}{119}\)
=>\(\dfrac{x-3}{113}+1+\dfrac{x-5}{115}+1=\dfrac{x-7}{117}+1+\dfrac{x-9}{119}+1\)
=>\(\dfrac{x+110}{113}+\dfrac{x+110}{115}=\dfrac{x+110}{117}+\dfrac{x+110}{119}\)
=>(x+110)(\(\dfrac{1}{113}+\dfrac{1}{115}-\dfrac{1}{117}-\dfrac{1}{119}\))=0
vì 1/113+1/115-1/117-1/119 khác 0
=>x+110=0=>x=-110
\(\dfrac{x-291}{1700}+\dfrac{x-293}{1698}+\dfrac{x-295}{1696}+\dfrac{x-297}{1694}=4\)
=>\(\dfrac{x-291}{1700}-1+\dfrac{x-293}{1698}-1+\dfrac{x-295}{1696}-1+\dfrac{x-297}{1694}-1+4=4\)=>\(\dfrac{x-1991}{1700}+\dfrac{x-1991}{1698}+\dfrac{x-1991}{1696}+\dfrac{x-1991}{1694}=0\)
=>(x-1991)(1/1700+1/1698+1/1696+1/1694)=0
vì 1/1700+1/1698+1/1696+1/1694 khác 0
=>x-1991=0=>x=1991