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TG
1
6 tháng 6 2016
Ta có: \(4x^2-21x+23+2\sqrt{x+1}=0\)
\(\Rightarrow-\left(x+1\right)+2\sqrt{x+1}+4x^2-20x+24=0\)
Đặt a = \(\sqrt{x+1}\left(a\ge0\right)\) , ta được pt: -a2 + 2a + 4x2 - 20x + 24 = 0
Có: \(\Delta'=1+4x^2-20x+24=4x^2-20x+25=\left(2x-5\right)^2\Rightarrow\sqrt{\Delta}=2x-5\)
\(\Rightarrow\orbr{\begin{cases}a=6-2x\\a=2x-4\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x+1}=6-2x\\\sqrt{x+1}=2x-4\end{cases}}\Rightarrow\orbr{\begin{cases}x+1=\left(6-2x\right)^2\\x+1=\left(2x-4\right)^2\end{cases}}}\)
Tới đây bạn tự giải
DG
1
UK
10 tháng 10 2017
Tự đặt điều kiện :v
\(\Leftrightarrow x^2\sqrt{x^2-4}+2x=0\)
Đặt \(\left(x;\sqrt{x^2-4}\right)=\left(a;b\right)\)
Phương trình đã cho tương đương với hệ
\(\left\{{}\begin{matrix}a^2b+2a=0\\b^2+4=a^2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a\left(ab+2\right)=0\\a^2-b^2=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left[{}\begin{matrix}a=0\\ab+2=0\end{matrix}\right.\\a^2-b^2=4\end{matrix}\right.\)
Tự giải tiếp các TH
11 tháng 10 2017
bạn giúp mk làm câu này được ko cấu trên mk ghi sai đề .
\(x+\dfrac{2x}{\sqrt{x^2-4}}=3\sqrt{5}\) 
DQ
4
10 tháng 9 2016
a/ PT <=> (x2 - 6x + 9) + (x - \(\sqrt{3x}\)) + (3 - \(\sqrt{3x}\)) = 0
<=> (\(\sqrt{x}-\sqrt{3}\))(\(\sqrt{3}x+x\sqrt{x}-3\sqrt{x}-3\sqrt{3}\)) + √x(\(\sqrt{x}-\sqrt{3}\)) + \(\sqrt{3}\left(\sqrt{3}-\sqrt{x}\right)\)= 0
<=> x = 3
1 tháng 4 2017
I Don't No
~~ tk nha ~`
─(♥)(♥)(♥)────(♥)(♥)(♥) __ ɪƒ ƴσυ’ʀє αʟσηє,
──(♥)██████(♥)(♥)██████(♥) ɪ’ʟʟ ɓє ƴσυʀ ѕɧα∂σѡ.
─(♥)████████(♥)████████(♥) ɪƒ ƴσυ ѡαηт тσ cʀƴ,
─(♥)██████████████████(♥) ɪ’ʟʟ ɓє ƴσυʀ ѕɧσυʟ∂єʀ.
──(♥)████████████████(♥) ɪƒ ƴσυ ѡαηт α ɧυɢ,
────(♥)████████████(♥) __ ɪ’ʟʟ ɓє ƴσυʀ ρɪʟʟσѡ.
──────(♥)████████(♥) ɪƒ ƴσυ ηєє∂ тσ ɓє ɧαρρƴ,
────────(♥)████(♥) __ ɪ’ʟʟ ɓє ƴσυʀ ѕɱɪʟє.
─────────(♥)██(♥) ɓυт αηƴтɪɱє ƴσυ ηєє∂ α ƒʀɪєη∂,
───────────(♥) __ ɪ’ʟʟ ʝυѕт ɓє ɱє.
NT
1
25 tháng 7 2018
Đặt \(\sqrt[3]{x}=a\). Ta có:
4a\(^2\)+21a+27=0giải phương trình bậc hai
26 tháng 6 2019
Of course she is right. Panties are more comfortable. Actually as a girl I don't have experience of men's boxers but few boys in my friend circle use panties and according to them that are more comfortable than their regular and bore boxers. One thing to keep in mind that use one size larger than your regular size , if you are using 90 or L size regularly then whenever you buy panty take 95 or XL in that way. Actually its told by one of my boy friend and its pracal also as woman inners are somewhat tight for men due to penis and testes.
As far as girls point of view concerned I think many men now using bikinis instead of frenchis. As bikinis provide more comfort to their private organs and provide some kind of pleasure and excitement as well .
Some of boys in my friend circle use panties. I had done their shopping with them on many occasions. My cousin sister's husband is also using bikinis since about last 6–7 years as regular innerwear and my she has no objection.
Basically its problem of thinking . We attach that small piece of cloth with gender ,if you see careful both men inners and women panties are very much similar in look only some design and fashion trends are introduced in panties to make them more attractive and sex appealing.
Only piece of cloth panty can't decide your gender. Its every one's individual matter what to wear and what not. As jeans, T shirts, sport wears, shoes, jerkins, sweat shirt and many more things are now common for boys and girls then why not panties? I think one should prefer his or her comfort rather than others opinion or objections.
L
26 tháng 6 2019
Of course she is right. Panties are more comfortable. Actually as a girl I don't have experience of men's boxers but few boys in my friend circle use panties and according to them that are more comfortable than their regular and bore boxers. One thing to keep in mind that use one size larger than your regular size , if you are using 90 or L size regularly then whenever you buy panty take 95 or XL in that way. Actually its told by one of my boy friend and its pracal also as woman inners are somewhat tight for men due to penis and testes.
As far as girls point of view concerned I think many men now using bikinis instead of frenchis. As bikinis provide more comfort to their private organs and provide some kind of pleasure and excitement as well .
Some of boys in my friend circle use panties. I had done their shopping with them on many occasions. My cousin sister's husband is also using bikinis since about last 6–7 years as regular innerwear and my she has no objection.
Basically its problem of thinking . We attach that small piece of cloth with gender ,if you see careful both men inners and women panties are very much similar in look only some design and fashion trends are introduced in panties to make them more attractive and sex appealing.
Only piece of cloth panty can't decide your gender. Its every one's individual matter what to wear and what not. As jeans, T shirts, sport wears, shoes, jerkins, sweat shirt and many more things are now common for boys and girls then why not panties? I think one should prefer his or her comfort rather than others opinion or objections.
29 tháng 4 2020
Bài 1:
Ta có: \(\left(2x^2+x-4\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x^2+x-4-2x+1\right)\left(2x^2+x-4+2x-1\right)=0\)
\(\Leftrightarrow\left(2x^2-x-3\right)\left(2x^2+3x-5\right)=0\)
\(\Leftrightarrow\left(2x^2+2x-3x-3\right)\left(2x^2-2x+5x-5\right)=0\)
\(\Leftrightarrow\left[2x\left(x+1\right)-3\left(x+1\right)\right]\left[2x\left(x-1\right)+5\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)\left(x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\\x-1=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=3\\x=1\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\\x=1\\x=\frac{-5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{3}{2};1;\frac{-5}{2}\right\}\)
\(4x^2-21x+23+2\sqrt{x+1}=0\) (x\(\ge-1\))
\(\Leftrightarrow\left(4x^2-20x+25\right)-\left(x+1+2\sqrt{x+1}+1\right)\)=0
\(\Leftrightarrow\left(2x-5\right)^2=\left(\sqrt{x+1}+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=\sqrt{x+1}+1\\2x-5=-\sqrt{x+1}-1\end{matrix}\right.\) ....
Ta có: \(4x^2-21x+23+2\sqrt{x+1}=0\left(Đkxđ:x\ge-1\right)\)
\(\Leftrightarrow4x^2-21x+23=-2\sqrt{x+1}\)
\(\Leftrightarrow16x^4+441x^2+529-168x^3+184x^2-966x=4\left(x+1\right)\)
\(\Leftrightarrow16x^4-168x^3+625x^2-970x+525=0\)
\(\Leftrightarrow\left(16x^3-120x^2+265x-175\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-\frac{5}{4}\right)\left(16x^2-100x+140\right)=0\)
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