Mở đầu về phương trình - Hỏi đáp

Mình sẽ trình bày lời giải thứ hai cho bài này:

Đặt x - 12 = a; 2x - 12 = b thì 24 - 3x = -(a+b). Như vậy, phương trình đã cho trở thành:

\(a^7+b^7-\left(a+b\right)^7=0\)

\(\Leftrightarrow\left(a+b\right)\left(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6\right)\)

\(-\left(a+b\right)\left(a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6\right)\)= 0.

\(\Leftrightarrow\left(a+b\right)\left(-7a^5b-14a^4b^2-21a^3b^3-14a^2b^4-7ab^5\right)=0\)

\(\Leftrightarrow\left(a+b\right).-7ab.\left(a^4+2a^3b+3a^2b^2+2ab^3+b^4\right)=0\)

\(\Leftrightarrow-7ab\left(a+b\right)\left[\left(a^2+ab\right)^2+a^2b^2+\left(b^2+ab\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\2x-12=0\\3x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=6\\x=8\end{matrix}\right.\)

Vậy nghiệm của phương trình là \(x=\left\{6;8;12\right\}\)

P/s: Đây là bài 1b, đề thi THPT chuyên Khoa học tự nhiên - Hà Nội, vòng 2 (Toán chuyên), 2020.

Lời giải:

Đặt $x-12=a; 2x-12=b; 24-3x=c$ thì $a+b=-c$

PT trở thành:

$a^7+b^7+c^7=0$

$\Leftrightarrow (a^3+b^3)(a^4+b^4)-a^3b^3(a+b)+c^7=0$

$\Leftrightarrow [(a+b)^3-3ab(a+b)][(a+b)^4+2a^2b^2-4ab(a+b)^2]-a^3b^3(a+b)+c^7=0$

$\Leftrightarrow (-c^3+3abc)(c^4+2a^2b^2-4abc^2)+a^3b^3c+c^7=0$

$\Leftrightarrow -c^7-2a^2b^2c^3+4abc^5+3abc^5+6a^3b^3c-12a^2b^2c^3+a^3b^3c+c^7=0$

$\Leftrightarrow -14a^2b^2c^3+7abc^5+7a^3b^3c=0$

$\Leftrightarrow abc(2abc^2-c^4-a^2b^2)=0$

Nếu $abc=0$ ta xét các TH:

$a=0\Rightarrow x=12$ (thỏa mãn)

$b=0\Rightarrow x=6$ (thỏa mãn )

$c=0\Rightarrow x=8$ (thỏa mãn)

Nếu $2abc^2-c^4-a^2b^2=0$

$\Leftrightarrow (c^2-ab)^2=0\Rightarrow ab=c^2$

$\Leftrightarrow ab=[-(a+b)^2]=(a+b)^2\Leftrightarrow a^2+ab+b^2=0$

$\Leftrightarrow (a+\frac{b}{2})^2+\frac{3}{4}b^2=0$

$\Rightarrow a+\frac{b}{2}=b=0\Rightarrow a=b=0$ (vô lý)

Vậy.......

a) ĐKXĐ : \(3x^2-4x\ne0\Leftrightarrow x\left(3x-4\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\frac{4}{3}\end{matrix}\right.\)

b) Rút gọn :

\(\frac{\left(3x\right)^2-4^2}{x\left(3x-4\right)}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)

Vậy : ...

c) Để \(\frac{3x+4}{x}=0\Leftrightarrow3x+4=0\Leftrightarrow x=-\frac{4}{3}\) ( thỏa mãn ĐKXĐ )

\(M=\left(4x^2-4x+1\right)+x+\frac{1}{x}+2019\)

\(M=\left(2x-1\right)^2+x+\frac{1}{x}+2019\)

Do x>0, áp dụng BĐT Cosi, ta có

\(x+\frac{1}{x}=\left(\frac{1}{x}+4x\right)-3x\ge4-3x\)

Mặt khác: \(\left(2x-1\right)\ge0\forall x\)

Dấu "=" xảy ra <=> x=1/2

<=> \(M\ge4-\frac{3}{2}+2019=...\)

+ Ta có: $3 < 4$

$\Rightarrow 2b + 3 < 2b + 4$ (cộng vào hai vế với $2b$)

+ Vậy ta có đpcm

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

Đặt \(\left\{{}\begin{matrix}x-3=a\\2x-3=b\end{matrix}\right.\)

\(\Rightarrow a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow a^3+b^3=a^3+b^3+3ab\left(a+b\right)\)

\(\Leftrightarrow ab\left(a+b\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-3=0\\3x-6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{3}{2}\\x=2\end{matrix}\right.\)

\(\left(x-3\right)^3+\left(2x-3\right)^3=27\left(x-2\right)^3\)

\(\Leftrightarrow\left(x-3+2x-3\right)\left[\left(x-3\right)^2-\left(x-3\right)\left(2x-3\right)+\left(2x-3\right)^2\right]=\left(3x-6\right)^3\)

\(\Leftrightarrow\left(3x-6\right)\left(x^2-6x+9-2x^2+9x-9+4x^2-12x+9\right)=\left(3x-6\right)^3\)

\(\Leftrightarrow\left(3x-6\right)\left(3x^2-9x+9\right)=\left(3x-6\right)\left(9x^2-36x+36\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-6=0\\3x^2-9x+9=9x^2-36x+36\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\6x^2-27x+27=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\3\left(2x-3\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=3\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là : \(S=\left\{2;\frac{3}{2};3\right\}\)

Ta có:

\(2\left(y^2+yz+z^2\right)+3x^2=36\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2zx+z^2\right)=36\)

\(\Leftrightarrow\left(x+y+z\right)^2=36-\left(x-y\right)^2-\left(x-z\right)^2\le36\)

\(\Leftrightarrow-6\le x+y+z\le6\)

\(\RightarrowĐPCM\)

Ta có: a+b+c=0

\(\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)

\(M=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ac}{a^2+c^2-b^2}\)

\(M=\frac{ab}{\left(a+b\right)^2-2ab-c^2}+\frac{bc}{\left(b+c\right)^2-2bc-a^2}+\frac{ac}{\left(a+c\right)^2-2ac-b^2}\)

\(M=\frac{ab}{\left(-c\right)^2-2ab-c^2}+\frac{bc}{\left(-a\right)^2-2bc-a^2}+\frac{ac}{\left(-b\right)^2-2ac-b^2}\)

\(M=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ac}{-2ac}\)

\(M=-\frac{1}{2}+-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)

Ta có: \(\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\right)\cdot\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\cdot\left(x-1\right)=\frac{3x}{5}-\frac{7}{15}\)

\(\Leftrightarrow\frac{14}{15}\cdot\left(x-1\right)=\frac{9x-7}{15}\)

\(\Leftrightarrow x-1=\frac{9x-7}{15}:\frac{14}{15}=\frac{9x-7}{14}\)

hay \(x=\frac{9x-7}{14}+1=\frac{9x-7}{14}+\frac{14}{14}=\frac{9x+7}{14}\)

\(\Leftrightarrow x\cdot14=9x+7\)

\(\Leftrightarrow14x-9x-7=0\)

\(\Leftrightarrow5x-7=0\)

\(\Leftrightarrow5x=7\)

hay \(x=\frac{7}{5}\)

Vậy: \(x=\frac{7}{5}\)

\(5+\dfrac{96}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\left(ĐKXĐ:x\ne\pm4\right)\)

\(\Leftrightarrow5+\dfrac{96}{\left(x-4\right)\left(x+4\right)}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)

\(\Leftrightarrow\dfrac{5\left(x-4\right)\left(x+4\right)+96}{\left(x-4\right)\left(x+4\right)}=\dfrac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)

\(\Rightarrow5\left(x-4\right)\left(x+4\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)

P/s: tới đây pn tự giải tiếp nha, kết quả \(x=8\) (nhận)

Giải phương trình

a, 5x(x-4)-5x² = 2 (11-x)

\(\Leftrightarrow5x^2-20x-5x^2=22-2x\)

\(\Leftrightarrow-18x=22\)

\(\Leftrightarrow x=\frac{-22}{18}\)

b, \(\frac{3}{x-3}-\frac{2}{x+3}=\frac{4x}{x^2-9}\left(x\ne\pm3\right)\)

\(\Leftrightarrow\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{4x}{x^2-9}\)

\(\Rightarrow3x+9-2x+6=4x\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=5\left(tm\right)\)

Kl: a,.........

b,.........

a)\(14-2x=0\)

\(\Leftrightarrow2x=14\)

\(\Leftrightarrow x=7\). Vậy \(S=\left\{7\right\}\)

b)\(3x-4=9+3x\)

\(\Leftrightarrow3x-3x=4+9\)

\(\Leftrightarrow0x=13\)(vô lí). Vậy \(S=\varnothing\)

c)\(3\left(x+1\right)=3x+3\)

\(\Leftrightarrow3\left(x+1\right)=3\left(x+1\right)\)

Vậy x thỏa mãn phương trình \(\forall x\)

có x²\(\ge\)0 với \(\forall\)x

x²-x+1=x²-2x\(\frac{1}{2}\)+\(\frac{1}{4}+\frac{3}{4}\)>0 với\(\forall x\)

\(\Rightarrow\frac{x^2}{x^2-x+1}\ge0\forall x\)

Dấu "="\(\Leftrightarrow x=0\)

\(5x^2-4xy+y^2-6x+8=0\)

\(\Leftrightarrow25x^2-20xy+5y^2-30x+40=0\)

\(\Leftrightarrow\left(5x-2y\right)^2+\left(y-15\right)^2=185=64+121=8^2+11^2\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}5x-2y=8\\y-15=11\end{matrix}\right.\\\left[{}\begin{matrix}5x-2y=11\\y-15=8\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}5x=2y+8\\y=26\end{matrix}\right.\\\left[{}\begin{matrix}5x=11+2y\\y=23\end{matrix}\right.\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}\left\{{}\begin{matrix}x=12\\y=26\end{matrix}\right.\left(thoaman\right)}\\\left\{{}\begin{matrix}x=11,4\\y=23\end{matrix}\right.\left(kothoaman\right)\end{matrix}\right.\)

Vậy S={26,12}

Giúp bạn câu 1 thôi (Mình lười lắm)

(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

Chúc bn học tốt!!

Câu 1:

Ta có : (a+b)⁴ = [(a+b)²]²

= (a² + 2ab + b²)²

= a⁴ + 4a²b² + b⁴ + 2a³b + a²b² + 2ab³

Câu 2:

a. Biểu thức A xác định khi \(\left[{}\begin{matrix}x\ne0\\x-3\ne0\\x^2-3x\ne0\end{matrix}\right.\)⇒ \(\left[{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)

Vậy ĐKXĐ của A là \(x\ne0;x\ne3\)

b. A = \(\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right)\div\frac{2x-2}{x}\)

= \(\left(\frac{\left(x-3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x.x}{\left(x-3\right)x}+\frac{9}{x\left(x-3\right)}\right)\div\frac{2x-2}{x}\)

= \(\frac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\div\frac{2\left(x-1\right)}{x}\)

= \(\frac{-6x+18}{x\left(x-3\right)}\times\frac{x}{2\left(x-1\right)}\)

= \(\frac{-6\left(x-3\right)}{x-3}\times\frac{1}{2\left(x-1\right)}\)

=\(-3\times\frac{1}{x-1}\)

= \(\frac{-3}{x-1}\)

Vậy A= \(\frac{-3}{x-1}\)

c. Để A nguyên hay \(\frac{-3}{x-1}\) có giá trị nguyên

thì x - 1 ϵ Ư(-3) = {-3; 3; -1; 1}

⇔ x ϵ {-2; 4; 0; 2}

Vậy A có giá trị nguyên khi x ϵ {-2; 4; 0; 2}

Học tốt nha ~