\(x^2+4x=23-10\sqrt{2}\Leftrightarrow x^2+4x+4=\left(x+2\right)^2=27-10\sqrt{2}=25-10\sqrt{2}+2=5^2-5.2\sqrt{2}+\left(\sqrt{2}\right)^2=\left(5-\sqrt{2}\right)^2=\left(\sqrt{2}-5\right)^2\Leftrightarrow\left[{}\begin{matrix}x+2=5-\sqrt{2}\\x+2=\sqrt{2}-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3-\sqrt{2}\\x=\sqrt{2}-7\end{matrix}\right.\)

\(b,+,x>\frac{1}{2}\Rightarrow2x>1\Rightarrow1-2x< 0\Rightarrow\left|1-2x\right|=-\left(1-2x\right)=2x-1\Rightarrow\left(2x-1\right)^2=\left(2x-1\right)\Leftrightarrow\left(2x-1\right)\left(2x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(loai\right)\\x=1\left(thoaman\right)\end{matrix}\right.\)\(+,x\le\frac{1}{2}\Rightarrow2x\le1\Rightarrow1-2x\ge0\Rightarrow\left|1-2x\right|=1-2x\Rightarrow\left(2x-1\right)^2=1-2x\Leftrightarrow2x\left(2x-1\right)=0\Leftrightarrow x\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(thoaman\right)\\x=\frac{1}{2}\left(thoaman\right)\end{matrix}\right..\)

\(c,Taco:\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(2x+1\right)^2\ge0\end{matrix}\right.mà:\left(x-2\right)^2+\left(2x+1\right)^2=0nên:\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(2x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-\frac{1}{2}\end{matrix}\right.\left(voli\right).Nên:x\in\varnothing\)

cảm ơn rất nhìu

a:

\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)

\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)

\(\Leftrightarrow30x+15-100-6x+4=24x-8\)\(\Leftrightarrow30x-6x-24x=100-4-8\)

\(\Leftrightarrow0x=88\)

Vậy pt vô nghiệm

b:

\(\dfrac{x-15}{23}+\dfrac{x-23}{15}-2=0\)

\(\Leftrightarrow\dfrac{x-15}{23}+\dfrac{x-23}{15}=2\)

\(\Leftrightarrow\dfrac{x-15}{23}-1+\dfrac{x-23}{15}-1=2-2\)

\(\Leftrightarrow\dfrac{x-15-23}{23}+\dfrac{x-23-15}{15}=0\)

\(\Leftrightarrow\dfrac{x-38}{23}+\dfrac{x-23}{15}=0\)

\(\Leftrightarrow\left(x+38\right)\left(\dfrac{1}{23}+\dfrac{1}{15}\right)=0\)

Vì \(\dfrac{1}{23}+\dfrac{1}{15}\ne0\) nên x + 38 =0 \(\Leftrightarrow x=-38\)

Vậy tập nghiện của pt S= {-38}

c:

\(\dfrac{3\left(2x+1\right)}{4}-\dfrac{5x+3}{6}+\dfrac{x+1}{3}=x+\dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\dfrac{12x+7}{12}\)

\(\Leftrightarrow9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)=12x+7\)

\(\Leftrightarrow18x+9-10x-6+4x+4=12x+7\)

\(\Leftrightarrow18x-10x+4x-12x=7-9+6-4\)

\(\Leftrightarrow0x=0\)

Vậy pt vô số nghiệm

1 số gợi ý

hpt \(\Leftrightarrow\left\{{}\begin{matrix}2x\left(2x-2y-1\right)=6\left(y+2\right)\\6y+12\sqrt{2x-1}=2y^2-2x+46\end{matrix}\right.\)(1)

Đặt \(\sqrt{2x-1}=t\left(t\ge0\right)\)

(1)\(\Leftrightarrow\left\{{}\begin{matrix}\left(t^2+1\right)\left(t^2-2y\right)=6\left(y+2\right)\left(2\right)\\6y+12t=2y^2-t^2+45\end{matrix}\right.\)

(2)\(\Leftrightarrow\left(t^2+4\right)\left(t^2-2y-3\right)=0\)

\(\Leftrightarrow t^2-2y-3=0\)

ta có hpt mới sau : \(\left\{{}\begin{matrix}t^2-2y-3=0\\2y^2-t^2+45=6y+12t\end{matrix}\right.\)

một cách trâu bò nhưng hiệu quả là

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{t^2-3}{2}\\2y^2-t^2-6y-12t+45=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{t^2-3}{2}\\2\left(\dfrac{t^2-3}{2}\right)^2-t^2-6\left(\dfrac{t^2-3}{2}\right)-12t+45=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{t^2-3}{2}\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\t=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=5\end{matrix}\right.\)

\(\left(a,b,n\in N\right)\left\{{}\begin{matrix}n^2=a+b\\n^3+2=a^2+b^2\end{matrix}\right.\)

Áp dụng BĐT cơ bản : \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\)

\(\rightarrow n^3+2=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\left(n^2\right)^2=\dfrac{1}{2}n^4\)

\(\Rightarrow n^3+2-\dfrac{n^4}{2}\ge0\)\(\Rightarrow0\le n\le2\)

Xét từng TH của n và kết quả nhận được là \(n=2\); (a,b) là hoán vị của (1,3)

what hell ?
Bạn giải hộ ai à?

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.vi diệu !

hok cũng giỏi ghê 

~ tự biên tự diễn hả ~

Sửa đề : \(x^2+3=..\) nhé

a. \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

\(\Leftrightarrow x=23\) (Vì \(\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)\ne0\) )

b. \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

\(\Leftrightarrow x=-100\) (Vì \(\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)\ne0\) )

ĐK: \(x\ge\frac{1}{2}\)

\(\hept{\begin{cases}x\left(2x-2y-1\right)=3\left(y+2\right)\left(1\right)\\3y+6\sqrt{2x-1}=y^2-x+23\left(2\right)\end{cases}}\)

pt (1) <=> \(2x^2-2xy-x-3y-6=0\)

<=> \(2x^2-x\left(2y+1\right)-\left(3y+6\right)=0\)

có \(\Delta=\left(2y+1\right)^2+4\left(3y+6\right)=4y^2+28y+49=\left(2y+7\right)^2\)

=> (1) có hai nghiệm: \(\orbr{\begin{cases}x_1=\frac{\left(2y+1\right)-\left(2y+7\right)}{4}=-\frac{3}{2}\left(loai\right)\\x_2=\frac{\left(2y+1\right)+\left(2y+7\right)}{4}=y+2\end{cases}}\)

+) Với \(x=y+2\) thế vào (2) ta có: 

\(3y+6\sqrt{2\left(y+2\right)-1}=y^2-\left(y+2\right)+23\)

<=> \(6\sqrt{2y+3}=y^2-4y+21\)

ĐK: \(y\ge-\frac{3}{2}\)

\(6\sqrt{2y+3}=y^2-4y+21\)

<=> \(6\sqrt{2y+3}-2y-12=y^2-6y+9\)

<=> \(\frac{2\left(9\left(2y+3\right)-\left(y+6\right)^2\right)}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)

<=> \(\frac{-2\left(y-3\right)^2}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)

<=> \(\left(y-3\right)^2\left(\frac{-2}{3\sqrt{2y+3}+y+6}-1\right)=0\)

<=> y - 3 = 0 

<=> y = 3 thỏa mãn 

khi đó x = y + 2 = 3 + 2 = 5 thỏa mãn

Kết luận:...