Bài 3: Phương trình đưa được về dạng ax + b = 0

Sai đề. Ví dụ: x = y = 1 => x² - 3xy + y² = 1² - 3.1.1 + 1² = -1

\(5\left(x+2\right)-8=7\left(2x-3\right)\\ \Leftrightarrow5x+10-8=14x-21\\\Leftrightarrow5x-14x=-21-10+8\\ \Leftrightarrow-9x=-23 \\ \Leftrightarrow x=\dfrac{23}{9}\)

Vậy tập nghiệm của phương trình \(S=\left\{\dfrac{23}{9}\right\}\)

\(\left(2x-1^2\right)-\left(2-x\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)-\left(2-x\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(1-2+x\right)\\\Leftrightarrow\left(2x-1\right)\left(-1+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\-1+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0,5\\x=1\end{matrix}\right. \)

Vậy tập nghiệm của phương trình \(S=\left\{0,5;1\right\}\)

\(6x\left(x+5\right)\ge3\left(2x^2+5\right)\\ \Leftrightarrow6x^2+30x\ge6x^2+15\\ \Leftrightarrow30x\ge15\\ \Leftrightarrow x\ge0,5\)

Vậy tập nghiệm của bất phương trình \(S=\left\{x|x\ge0,5\right\}\)

(Bất phương trình không biết trình bày tập nghiệm vì mới lớp 6)

5(x+2)-8=7(2x-3)

\(\Leftrightarrow\)5x+10-8=14x-21

\(\Leftrightarrow\)5x-14x=-10+8-21

\(\Leftrightarrow\)-9x=-23

\(\Leftrightarrow\)x=\(\dfrac{23}{9}\)

Vậy S={\(\dfrac{23}{9}\)}

(2x-1)²-(2-x)(2x-1)=0

\(\Leftrightarrow\)(2x-1)[(2x-1)-(2-x)]=0

\(\Leftrightarrow\)(2x-1)(2x-1-2+x)=0

\(\Leftrightarrow\)(2x-1)(3x-3)=0

\(\Leftrightarrow\)2x-1=0 hoặc 3x-3=0

\(\Leftrightarrow\)2x-1=0

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

\(\Leftrightarrow3x-3=0\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

Vậy S={\(\dfrac{1}{2}\);1}

6x(x+5)\(\ge\)3(2x²+5)

\(\Leftrightarrow\)6x²+30x\(\ge\)6x²+15

\(\Leftrightarrow\)30x\(\ge\)15

\(\Leftrightarrow\)x\(\ge\)\(\dfrac{1}{2}\)

Vậy S={x/x\(\ge\)\(\dfrac{1}{2}\)}

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-4x^3+2x^2-5x^3+10x^2-5x+2x^2-4x+2=0\)

\(\Leftrightarrow2x^2\left(x^2-2x+1\right)-5x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-5x+2\right)\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-x-4x+2\right)\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)^2\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)

\(\Leftrightarrow2x^3\cdot\left(x-1\right)-7x^2\cdot\left(x-1\right)+7x\cdot\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x^3-1\right)-7x\cdot\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x-1\right)\cdot\left(x^2+x+1\right)-7x\cdot\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[2\left(x^2+x+1\right)-7x\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2+2x+2-7x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-5x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-x-4x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[x\cdot\left(2x-1\right)-2\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\x-2=0\\2x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{1}{2};x_2=1;x_3=2\)

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)

\(\Leftrightarrow\)\(2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-4x^2-3x^2+6x+x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(2x^2-3x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(2x^2-2x-x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(2x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(x-1\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy ...................

\(x^3-x^2+x-1=0\)

\(\Leftrightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+1\right)=0\)

Dễ thấy: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+1\ge1>0\forall x\) (vô nghiệm)

Nên \(x-1=0\Rightarrow x=1\)

$x^3-x^2+x-1=0$

\(\Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+1\right)=0\)

Vì \(x^2+1>0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

S = {1}

\(x^4+5x^3-8x-40=0\)

\(\Leftrightarrow x^3\left(x+5\right)-8\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^3-8\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x+5\right)=0\)

Ta có : \(x^2+2x+4=x^2+2x+1+3=\left(x+1\right)^2+3\ge3\)\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

\(x^4+5x^3-8x-40=0\)

\(\Leftrightarrow x^4+3x^3-10x^2+2x^3+6x^2-20x+4x^2+12x-40=0\)

\(\Leftrightarrow x^2\left(x^2+3x-10\right)+2x\left(x^2+3x-10\right)+4\left(x^2+3x-10\right)=0\)

\(\Leftrightarrow\left(x^2+3x-10\right)\left(x^2+2x+4\right)=0\)

\(\Leftrightarrow\left(x^2-2x+5x-10\right)\left(x^2+2x+4\right)=0\)

\(\Leftrightarrow\left[x\left(x-2\right)+5\left(x-2\right)\right]\left(x^2+2x+4\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+5\right)\left(x^2+2x+4\right)=0\)

Dễ thấy: \(x^2+2x+4=x^2+2x+1+3=\left(x+1\right)^2+3>0\) (vô nghiệm)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

mình nhầm m=2 hoặc m=3

Vậy với m=2 hoặc m=3 thì phương trình đã cho vô nghiệm

Để phương trình đã cho vô nghiệm buộc \(a=0\Leftrightarrow m^2-5m+6=0\\ \Leftrightarrow\left(m-2\right)\left(m-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

PT Ax-B có vô số nghiệm chi khi A=0 và B =0

<=>

\(\left\{{}\begin{matrix}m^2-5m+6=0\\2m-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-2\right)\left(m-3\right)\\m=\dfrac{3}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m=2\\m=3\\m=\dfrac{3}{2}\end{matrix}\right.\) => vô nghiệm

Kết luận không có m thủa mãn

Bài 2:

a) Áp dụng BĐT AM - GM ta có:

\(\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{1}{4a}+\dfrac{1}{4b}\) \(\ge2\sqrt{\dfrac{1}{4^2ab}}=\dfrac{2}{4\sqrt{ab}}=\dfrac{1}{2\sqrt{ab}}\)

\(\ge\dfrac{1}{a+b}\) (Đpcm)

b) Trừ 1 vào từng vế của BĐT ta được BĐT tương đương:

\(\left(\frac{x}{2x+y+z}-1\right)+\left(\frac{y}{x+2y+z}-1\right)+\left(\frac{z}{x+y+2z}-1\right)\le\frac{-9}{4}\)

\(\Leftrightarrow-\left(x+y+z\right)\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\le-\frac{9}{4}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\ge\frac{9}{4}\)

Áp dụng BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) ta có:

\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)

\(\ge\dfrac{9}{2x+y+z+x+2y+z+x+y+2z}=\dfrac{9}{4\left(x+y+z\right)}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\ge\frac{9}{4}\)

\(\Leftrightarrow\dfrac{x}{2x+y+z}+\dfrac{y}{x+2y+z}+\dfrac{z}{x+y+2z}\le\dfrac{3}{4}\) (Đpcm)

Bài 1:

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT\ge\dfrac{\left(a+b\right)^2}{a-1+b-1}=\dfrac{\left(a+b\right)^2}{a+b-2}\)

Nên cần chứng minh \(\dfrac{\left(a+b\right)^2}{a+b-2}\ge8\)

\(\Leftrightarrow\left(a+b\right)^2\ge8\left(a+b-2\right)\)

\(\Leftrightarrow a^2+2ab+b^2\ge8a+8b-16\)

\(\Leftrightarrow\left(a+b-4\right)^2\ge0\) luôn đúng

a, x² -2xy+3y² -4y+2=0

\(\Leftrightarrow\)(x²-2xy+y²)+(y²-2y+1)+(y²-2y+1)=0

\(\Leftrightarrow\) (x-y)²+(y-1)²+(y-1)²=0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=y\\y=1\end{matrix}\right.\)\(\Leftrightarrow\) x=y=1

Gọi khối lượng cá bắt đc trong 1 ngày theo kế hoạch là x tấn ( x > 0)

Khối lượng cá bắt đc trong thực tế là x + 1 ( tấn )

Tổng khối lượng cá bắt đc theo kế hoạch là 8x ( tấn )

Tổng khối lượng cá bắt đc trong thực tế là 8x + 2 ( tấn )

Theo bài ta có phương trình sau:

\(\dfrac{8x}{x}-\dfrac{8x+2}{x+1}=1\)

\(\dfrac{8\left(x+1\right)-8x-2}{x+1}=\dfrac{x+1}{x+1}\)

\(8x+8-8x-2=x+1\)

\(-x=-5\Rightarrow x=5\)

Vậy Khối lượng cá bắt đc trong thực tế là x + 1 = 5 + 1 = 6 tấn.

Làm bừa :))