Phương trình bậc nhất một ẩn

ko xài pc dc k ?

Không 1 ai trả lời

Vì a;b dương nên \(a^3-b^3< a^3+b^3\)

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

\(\Rightarrow a^2+ab+b^2< 1\)

Lời giải:

a) $(x+3)(x-3)=16$

$\Leftrightarrow x^2-9=16$

$\Leftrightarrow x^2-25=0$

$\Leftrightarrow (x-5)(x+5)=0$

$\Rightarrow x-5=0$ hoặc $x+5=0$

$\Rightarrow x=5$ hoặc $x=-5$

b)

$x^2-9x+20=0$

$\Leftrightarrow (x^2-5x)-(4x-20)=0$

$\Leftrightarrow x(x-5)-4(x-5)=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Rightarrow x-4=0$ hoặc $x-5=0$

$\Rightarrow x=4$ hoặc $x=5$

c)

ĐK: $x\neq \pm 1$

PT $\Leftrightarrow \frac{2}{(x+1)^2}-\frac{5}{(x-1)^2}=\frac{3}{1-x^2}$

$\Leftrightarrow \frac{2(x-1)^2}{(x-1)^2(x+1)^2}-\frac{5(x+1)^2}{(x+1)^2(x-1)^2}=\frac{3(1-x^2)}{(x-1)^2(x+1)^2}$

$\Rightarrow 2(x-1)^2-5(x+1)^2=3(1-x^2)$

$\Leftrightarrow x=\frac{-3}{7}$ (tm)

d)

$-2x=|x^3-x|\geq 0$

$\Rightarrow x\leq 0$

Do đó: $|x^3-x|=|x(x^2-1)|=|x|.|x^2-1|=-x|x^2-1|$

PT trở thành:

$-x|x^2-1|=-2x$

$\Leftrightarrow -x(|x^2-1|-2)=0$

\(\Rightarrow \left[\begin{matrix} x=0\\ x^2-1=2\\ x^2-1=-2\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\pm \sqrt{3}\\ x^2=-1< 0(\text{vô lý})\end{matrix}\right.\)

Kết hợp với đk $x\leq 0$ suy ra $x=0$ hoặc $x=-\sqrt{3}$

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

ĐKXĐ : \(x\ne1;x\ne-1\)

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

\(\Rightarrow\left(x+1\right)\left(2x+1\right)=\left(x-1\right)\left(5x-5\right)\)

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

\(\Leftrightarrow-3x^2+13x-4\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ................

a: \(\dfrac{2}{\left(x+1\right)^2}-\dfrac{5}{\left(x-1\right)^2}=\dfrac{-3}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow2\left(x-1\right)^2-5\left(x+1\right)^2=-3\left(x^2-1\right)\)

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

=>-14x-6=0

hay x=-3/7

b: \(\left|x^2-x\right|=-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=4x^2\\x< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(3x+1\right)=0\\x< =0\end{matrix}\right.\Leftrightarrow x\in\left\{0;-\dfrac{1}{3}\right\}\)

|x-1|=5

Nếu x\(\ge\)1 thì x-1=5\(\Leftrightarrow\)x=6 (t/m)

Nếu x<1 thì x-1=-5\(\Leftrightarrow\)x=-4 (t/m)

Vậy x\(\in\){-4;6}

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

\(\Leftrightarrow4x^2+x-3=2x^2+7x+5\)

\(\Leftrightarrow2x^2-6x-8=0\)

\(\Leftrightarrow2x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow2\left(x+1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)

Vậy tập nghiệm của pt là:\(S=\left\{-1;4\right\}\)

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

\(\Rightarrow4x-3=2x+5\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)

Vậy x = 4

ĐKXĐ : \(x\ne3\)

\(\dfrac{1}{x-3}+2=\dfrac{x-5}{3-x}\)

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

=> \(1+2x-6=5-x\)

=> \(3x=10\)

=> \(x=\dfrac{10}{3}\) ( Thỏa mãn ĐKXĐ )

Vậy ...

bài này dễ mà :(

<=> \(\dfrac{1}{x-3}+\dfrac{2\left(x-3\right)}{x-3}+\dfrac{x-5}{x-3}=0\)

ĐK: x \(\ne3\)

\(\dfrac{1+2x-6+x-5}{x-3}=0\)

\(3x-10=0\)

=> x=\(\dfrac{10}{3}\)

x³-3x²-2x+6=0(Leftrightarrow)x²(x-3)-2(x-3)=0

(Leftrightarrow)(x²-2)(x-3)=0

Vì x²-2<0 nên x-3=0

(Leftrightarrow)x=3

\(x^3-3x^2-2x+6=0\)

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

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

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

\(\Leftrightarrow x-3=0\) hoặc \(x^2-2=0\)

+)Nếu x - 3 = 0 \(\Leftrightarrow\) x = 3

+)Nếu x² - 2 = 0 \(\Leftrightarrow\) x² = 3 \(\Leftrightarrow\) x = \(\pm\sqrt{2}\)

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

Chúc bạn học tốt!