Hãy thử lại và cho viết các khẳng định sau đây có đúng không ?
a) \(x^3+3x=2x^2-3x+1\Leftrightarrow x=-1\)
b) \(\left(z-2\right)\left(z^2+1\right)=2z+5\Leftrightarrow z=3\)
\(\hept{\begin{cases}3x^2+2y+1=2z\left(x+2\right)\\3y^2+2z+1=2x\left(y+2\right)\\3z^2+2x+1=2y\left(z+2\right)\end{cases}\Leftrightarrow\hept{\begin{cases}3x^2+2y+1=2xz+4z\\3y^2+2z+1=2xy+4x\\3z^2+2x+1=2yz+4y\end{cases}}}\)
Cộng 3 vế vào rồi chuyển vế ta được
\(2x^2+2y^2+2z^2-2xy-2yz-2zx+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2 +\left(z-x\right)^2+\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)
Dễ thấy VP > 0
Dấu "=" khi x = y = z = -1
\(\left(x^2-x+1\right)^4-6x^2\left(x^2-x+1\right)^2+5x^4=0\)
\(\Leftrightarrow\left[\left(x^2-x+1\right)^2\right]^2-2\left(x^2-x+1\right)^2.3x^2+\left(3x^2\right)^2-4x^4=0\)
\(\Leftrightarrow\left[\left(x^2-x+1\right)^2-3x^2\right]^2-\left(2x^2\right)^2=0\)
\(\Leftrightarrow\left[\left(x^2-x+1\right)^2-3x^2+2x^2\right]\left[\left(x^2-x+1\right)^2-3x^2-2x^2\right]=0\)
\(\Leftrightarrow\left[\left(x^2-x+1\right)^2-x^2\right]\left[\left(x^2-x+1\right)^2-5x^2\right]=0\)
\(\Leftrightarrow\left(x^2-x+1+x^2\right)\left(x^2-x+1-x^2\right)\left(x^4-2x^3-4x^2+1\right)=0\)
\(\Leftrightarrow\left(2x^2-x+1\right)\left(1-x\right)\left(x+1\right)\left(x^3-2x^2-x+1\right)=0\)
Mấy bạn cho mình gửi tạm nha, xíu mình nhờ CTV xóa :(
Hãy thử lại và cho biết các khẳng định sau có đúng không: z – 2 z 2 + 1 = 2 z + 5 ⇔ z = 3
z – 2 z 2 + 1 = 2 z + 5 ⇔ z = 3
Kết luận này sai vì thay z = 3 vào phương trình ta được:
VT = 3 - 2 3 2 + 1 = 1 . 9 + 1 = 10
VP = 2.3 + 5 = 6 + 5 = 11
⇒ VT ≠ VP
Tìm x,biết:
a/\(x+5x^2=0\Leftrightarrow......\)
b/\(x+1=\left(x+1\right)^2\Leftrightarrow..........\)
c/\(x^3+x=0\Leftrightarrow.......\)
d/\(5x\left(x-2\right)-\left(2-x\right)=0\)
e/\(x\left(2x-1\right)+\frac{1}{3}-\frac{2}{3}x=0\Leftrightarrow........\)
g/\(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow.....\)
h/\(x^2-3x=0\Leftrightarrow.....\)
i/\(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow.....\)
Tìm x,biết:
a/
\(\Leftrightarrow\) x = 0 hoặc 1 + 5x = 0
1) x = 0
2) 1+ 5x = 0 \(\Leftrightarrow\) x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{0;\frac{-1}{5}\right\}\)
b/
\(\Leftrightarrow\) (x+1) - (x+1)2 = 0
\(\Leftrightarrow\) ( x+ 1)(1-x-1) = 0
\(\Leftrightarrow\) (x+1).(-x) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x = 0
\(\Leftrightarrow\) x= -1 ; 0
Vậy: S=\(\left\{-1;0\right\}\)
c/
\(\Leftrightarrow\) x(x2 + 1) = 0
\(\Leftrightarrow\) x = 0 hoặc x2 + 1 = 0
Ta có : x2 + 1 \(\ge\) 0 vs mọi x
Vậy: S = \(\left\{0\right\}\)
d/0
\(\Leftrightarrow\) 5x(x-2) + (x - 2) = 0
\(\Leftrightarrow\) (x - 2)(5x+1) = 0
\(\Leftrightarrow\) x - 2 = 0 hoặc 5x+ 1 = 0
\(\Leftrightarrow\) x = 2 hoặc x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{\frac{-1}{5};2\right\}\)
g/
x = 4 hoặc x = 2
Vậy: S= \(\left\{2;4\right\}\)
h/
\(\Leftrightarrow\) x = 0 hoặc x = 3
Vậy: S = \(\left\{0;3\right\}\)
Vậy: S= \(\left\{0;3\right\}\)
i/
4x(x+1)-8(x+1) = 0
\(\Leftrightarrow\) 4(x+1) (x - 2) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x= -1 hoặc x = 2
Vậy: S=\(\left\{-1;2\right\}\)
ta có:(vế phải)2\(\le3\left(\frac{x^3}{y+z}+\frac{y^3}{z+x}+\frac{z^3}{x+y}\right)\)
cần chứng minh:
(vế trái)2/3\(\ge\frac{x^3}{y+z}+\frac{y^3}{z+x}+\frac{z^3}{x+y}\)
\(\Leftrightarrow\frac{x}{y+z}\left(\frac{x^3+\frac{1}{3}}{y+z}-x^2\right)+...\ge0\)
\(\Leftrightarrow\frac{x^2}{y+z}\left(x-y\right)\left(x-z\right)+\frac{y^2}{z+x}\left(y-x\right)\left(y-z\right)+\frac{z^2}{x+y}\left(z-x\right)\left(z-y\right)\ge0\)
bđt luôn đúng vì là bđt schur mở rộng
Rút gọn các biểu thức :
a, \(\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)\)
b, \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(c,\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\)
\(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6
\(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6
\(\Leftrightarrow\) 11x = 9
\(\Leftrightarrow\) x = 0,8
Vậy S = {0,8}
2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12
\(\Leftrightarrow\) \(\frac{6.\left(x+1\right)}{12}-\frac{x+3}{12}=\frac{12.3}{12}-\frac{4.\left(5-3x\right)}{12}\)
\(\Leftrightarrow\) 6x + 6 - x + 3 = 36 - 20 - 12x
\(\Leftrightarrow\) 6x - x + 12x = -6 - 3 + 36 - 20
\(\Leftrightarrow\) 17x = 7
\(\Leftrightarrow\) x = \(\frac{7}{17}\)
Vậy S = {\(\frac{7}{17}\)}
3) x - \(\frac{x+1}{3}\) = \(\frac{2x-1}{5}\) Mc : 15
\(\Leftrightarrow\) \(\frac{15.x}{15}-\frac{5.\left(x+1\right)}{15}=\frac{3.\left(2x-1\right)}{15}\)
\(\Leftrightarrow\) 15x - 5x - 5 = 6x - 3
\(\Leftrightarrow\) 15x - 5x - 6x = 5 - 3
\(\Leftrightarrow\) 4x = 2
\(\Leftrightarrow\) x = \(\frac{2}{4}=\frac{1}{2}\)
Vậy S = {\(\frac{1}{2}\)}
4) \(\frac{2x+7}{3}-\frac{x-2}{4}=-2\) Mc : 12
\(\Leftrightarrow\) \(\frac{4.\left(2x+7\right)}{12}-\frac{3.\left(x-2\right)}{12}=\frac{12.\left(-2\right)}{12}\)
\(\Leftrightarrow\) 8x + 28 -3x + 6 = -24
\(\Leftrightarrow\) 8x - 3x = -28 - 6 -24
\(\Leftrightarrow\) 5x = -58
\(\Leftrightarrow\) x = -11,6
Vậy S = {-11,6}
5) \(\frac{2x-3}{4}-\frac{4x-5}{3}=\frac{5-x}{6}\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(2x-3\right)}{12}-\frac{4.\left(4x-5\right)}{12}=\frac{2.\left(5-x\right)}{12}\)
\(\Leftrightarrow\) 6x - 9 - 16x + 20 = 10 - 2x
\(\Leftrightarrow\) 6x - 16x + 2x = 9 - 20 + 10
\(\Leftrightarrow\) -8x = -1
\(\Leftrightarrow\) x = \(\frac{1}{8}\)
Vậy S = {\(\frac{1}{8}\)}
6) \(\frac{12x+1}{4}=\frac{9x+1}{3}-\frac{3-5x}{12}\) Mc : 12
\(\Leftrightarrow\frac{3.\left(12x+1\right)}{12}=\frac{4.\left(9x+1\right)}{12}-\frac{3-5x}{12}\)
\(\Leftrightarrow\) 36x + 3 = 36x + 4 - 3 + 5x
\(\Leftrightarrow\) 36x - 36x - 5x = -3 + 4 - 3
\(\Leftrightarrow\) -5x = -2
\(\Leftrightarrow x=\frac{2}{5}\)
7) \(\frac{x+6}{4}\) - \(\frac{x-2}{6}-\frac{x+1}{3}=0\) Mc : 12
\(\Leftrightarrow\) \(\frac{3.\left(x+6\right)}{12}-\frac{2.\left(x-2\right)}{12}-\frac{4.\left(x+1\right)}{12}=0\)
\(\Leftrightarrow\) 3x + 18 - 2x + 4 - 4x - 4 = 0
\(\Leftrightarrow\) 3x - 2x - 4x = -18 - 4 + 4
\(\Leftrightarrow\) -3x = -18
\(\Leftrightarrow\) x = 6
Vậy S = {6}
8) x\(^2\) - x - 6 = 0
\(\Leftrightarrow\) x\(^2\) + 2x - 3x - 6 = 0
\(\Leftrightarrow\) x.(x + 2) - 3.(x + 2) = 0
\(\Leftrightarrow\) (x - 3).(x + 2) = 0
\(\Leftrightarrow\) x - 3 = 0 hoặc x + 2 = 0
\(\Leftrightarrow\) x = 3 hoặc x = -2
Vậy S = {3; -2}
Điều kiện: $ - \frac{1}{3} \le x \le 6$
Ta nhẩm thấy x = 5 là nghiệm của PT, thêm bớt và trục căn thức ta có:
Phương trình $ \Leftrightarrow \left( {\sqrt {3x + 1} - 4} \right) - \left( {\sqrt {6 - x} - 1} \right) + \left( {3{x^2} - 14x - 5} \right) = 0$
$ \Leftrightarrow \frac{{3\left( {x - 5} \right)}}{{\sqrt {3x + 1} + 4}} + \frac{{x - 5}}{{\sqrt {6 - x} + 1}} + \left( {3x + 1} \right)\left( {x - 5} \right) = 0$
$ \Leftrightarrow \left( {x - 5} \right)\left[ {\frac{3}{{\sqrt {3x + 1} + 4}} + \frac{1}{{\sqrt {6 - x} + 1}} + \left( {3x + 1} \right)} \right] = 0 \Leftrightarrow \left( {x - 5} \right)g\left( x \right) = 0$
Với điều kiện trên ta thấy g(x) > 0 vậy x = 5 là nghiệm của PT.
theo định lí đi dép tổ ong thì 2 trong 3 số x-2;y-2;z-2 cùng dấu
giả sử \(\left(x-2\right)\left(y-2\right)\ge0\Leftrightarrow xy-2\left(x+y\right)+4\ge0\)
\(\Leftrightarrow xy-2\left(6-z\right)+4\ge0\)
<=>xy-8+2z>(=)0
<=>xyz+2z^2-8z>(=)0
<=>xyz>(=)8z-2z^2
\(x^2-xy+y^2\ge\frac{x^2+y^2}{2}\ge\frac{\left(x+y\right)^2}{4}=\frac{\left(6-z\right)^2}{4}=\frac{z^2}{4}-3z+9\)
xz+yz=z(x+y)=x(6-z)=6z-z2
\(\Rightarrow x^2+y^2+z^2-xy-yz-zx+xyz\ge\frac{z^2}{4}-3z+9+z^2+z^2-6z+8z-z^2=\frac{z^2}{4}-z+9=\left(\frac{z}{2}-1\right)^2+8\ge8\)