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WTF đăng một loạt vầy ai dám làm @@
Mấy bài này trong sách bài tập cx có bài mẫu
tự lật sách ra học ik , đăng 1 loạt ai giải cho chép zô hết
\(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
Đặt \(x^2+5x=a\)
=> \(\left(a-6\right)\left(a+6\right)=a^2-36\ge-36\)
\(x\left(x+5\right)=0\) thì biểu thức nhỏ nhất
<=> x = 0 hoặc x = -5
1. \(x\left(y-4\right)=35-5\left(y-4\right)\) với y= 4 không phải nghiệm y khác 4
\(x=\frac{35}{y-4}-1\)
y=4+35/n
x=n-1
\(\hept{\begin{cases}n=\left\{-7,-5,-1,1,5,7\right\}\\y=\left\{-1,-3,-31,39,11,9\right\}\\x=n-1=\left\{-8,-6,-2,0,4,6\right\}\end{cases}}\)
2.x^2+x+6=y^2
4x^2+4x+1=4y^2-23
(2x+1)^2=4y^2-23
=>4y^2-23=t^2
(2y)^2-t^2=23
=>\(\hept{\begin{cases}y=+-6\\t=+-11\end{cases}\Rightarrow\hept{\begin{cases}2x+1=11\\2x+1=-11\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\x=-6\end{cases}}}\)
để \(\dfrac{x^3+x^2-x-1}{x^3+2x-3}=0\) thì
x3+x2-x-1=0
=>(x3+x2)-(x+1)=0
=>x2(x+1)-(x+1)=0
=>(x+1)(x2-1)=0
=>(x+1)(x-1)(x+1)=0
=>(x+1)2(x-1)=0
=>\(\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
vậy x=-1 hoặc x=1
\(A=\frac{\left|x-1\right|+\left|x\right|-x}{3x^2+4x+1}=\frac{1-x-x-x}{3x^2+3x+x+1}=\frac{1-3x}{\left(x+1\right)\left(3x+1\right)}\)
\(B=\frac{\left|2x-1\right|+x}{3x^2-22x+7}=\frac{1-2x+x}{3x^2-21x-x+7}=\frac{1-x}{\left(x-7\right)\left(3x-1\right)}\)
1.
<=> 7 - 2x - 4 = -x - 4
<=> -2x + x = -4 -7 + 4
<=> -x = -7
<=> x = 7
Vậy S = { 7 }
2.
<=> \(\frac{2\left(3x-1\right)}{6}\)= \(\frac{3\left(2-x\right)}{6}\)
<=> 2( 3x - 1 ) = 3( 2 - x )
<=> 6x -2 = 6 - 3x
<=> 6x + 3x = 6 + 2
<=> 9x = 8
<=> x = \(\frac{8}{9}\)
Vậy S = \(\left\{\frac{8}{9}\right\}\)
3.
<=> \(\frac{6x+10}{3}-\frac{x}{2}=5-\frac{3x+3}{4}\)
<=> \(\frac{4\left(6x+10\right)}{12}-\frac{6x}{12}=\frac{60}{12}-\frac{3\left(3x+3\right)}{12}\)
<=> 4( 6x + 10 ) - 6x = 60 - 3( 3x + 3 )
<=> 24x + 40 - 6x = 60 - 9x -9
<=> 18x + 40 = 51 - 9x
<=> 18x + 9x = 51 - 40
<=> 27x = 11
<=> x = \(\frac{11}{27}\)
Vậy S = \(\left\{\frac{11}{27}\right\}\)
<=>
Khó phết chứ chả đùa
Bài 1:
1.Đặt \(A=x^2+y^2-3x+2y+3\)
\(=x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+y^2+2y+1+2\)
\(=\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2-\frac{9}{4}+2\)
\(=\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2-\frac{1}{4}\)
Vì \(\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0;\forall x\\\left(y+1\right)^2\ge0;\forall y\end{cases}}\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2\ge0;\forall x,y\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\left(y+1\right)^2-\frac{1}{4}\ge0-\frac{1}{4};\forall x,y\)
Hay \(A\ge\frac{-1}{4};\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-1\end{cases}}\)
VẬY MIN A=\(\frac{-1}{4}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-1\end{cases}}\)
Bài 2:
1.Đặt \(C=12x-4x^2+3\)
\(=-4x^2+12x+3\)
\(-4x^2+12x-9+12\)
\(=-\left(4x^2-12x+9\right)+12\)
\(=-\left(2x-3\right)^2+12\)
Vì \(-\left(2x-3\right)^2\le0;\forall x\)
\(\Rightarrow-\left(2x-3\right)^2+12\le0+12;\forall x\)
Hay \(C\le12;\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x-3\right)^2=0\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy MAX C=12 \(\Leftrightarrow x=\frac{3}{2}\)
2. Đặt \(D=12x-8y-4x^2-y^2-2\)
\(=-4x^2-y^2-8y+12x-2\)
\(=-\left(4x^2+y^2+8y-12x+2\right)\)
\(=-\left[\left(4x^2-12x+9\right)+\left(y^2+8y+16\right)-23\right]\)
\(=-\left(2x-3\right)^2-\left(y+4\right)^2+23\)
Vì \(\hept{\begin{cases}-\left(2x-3\right)^2\le0;\forall x\\-\left(y+4\right)^2\le0;\forall y\end{cases}}\)
\(\Rightarrow-\left(2x-3\right)^2-\left(y+4\right)^2\le0;\forall x,y\)
\(\Rightarrow-\left(2x-3\right)^2-\left(y+4\right)^2+23\le0+23;\forall x,y\)
Hay \(D\le23;\forall x,y\)
Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}\left(2x-3\right)^2=0\\\left(y+4\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-4\end{cases}}\)
Vậy MAX D= 23 \(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-4\end{cases}}\)
Bài 1:
2. Đặt \(B=2x^2+5y^2+4xy+8x-4y+15\)
\(=\left(x^2+4xy+4y^2\right)+\left(y^2-4y+4\right)+\left(x^2+8x+16\right)-5\)
\(=\left(x+2y\right)^2+\left(y-2\right)^2+\left(x+4\right)^2-5\)
Vì \(\hept{\begin{cases}\left(x+2y\right)^2\ge0;\forall x,y\\\left(y-2\right)^2\ge0;\forall x,y\\\left(x+4\right)^2\ge0;\forall x,y\end{cases}}\)
\(\Rightarrow\left(x+2y\right)^2+\left(y-2\right)^2+\left(x+4\right)^2\ge0;\forall x,y\)
\(\Rightarrow\left(x+2y\right)^2+\left(y-2\right)^2+\left(x+4\right)^2-5\ge0-5;\forall x,y\)
Hay \(B\ge-5;\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+2y\right)^2=0\\\left(y-2\right)^2=0\\\left(x+4\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=2\\x=-4\end{cases}}\)
Vậy MIN \(B=-5\)\(\Leftrightarrow\hept{\begin{cases}y=2\\x=-4\end{cases}}\)