Ôn tập: Phương trình bâc nhất một ẩn

\(\left(m-2\right)x=3\\ \Leftrightarrow x=\dfrac{3}{m-2}\left(m\ne2\right)\)

Để pt có nghiệm nguyên âm thì \(m-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\) và m-2 âm

\(\Rightarrow m-2\in\left\{-3;-1\right\}\Rightarrow m\in\left\{-1;1\right\}\)

\(a,4x^2-1-x\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-x\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\\ b,ĐKXĐ:x\ne\pm3\\ \dfrac{x+3}{x-3}+\dfrac{x-3}{x+3}=\dfrac{x^2+2x}{x^2-9}+1\\ \Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+2x}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2+2x}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2-9}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+6x+9+x^2-6x+9-x^2-2x-x^2+9}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow-2x+27=0\\ \Leftrightarrow x=\dfrac{27}{2}\left(tm\right)\)

=>7(x-3)-4(2x+5)-14(x-1)=28

=>7x-21-8x-20-14x+14=28

=>-15x-27=28

=>-15x=55

hay x=-11/3

sai đề rồi b

\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)

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

a, \(x^2+7x+10-12x+9=x^2-10x+25\)

\(\Leftrightarrow5x=6\Leftrightarrow x=\dfrac{6}{5}\)

b, bạn ktra lại đề nhé 

c, \(x^2-4+3x+3=3+x^2-x-2\)

\(\Leftrightarrow x^2+3x-1=x^2-x+1\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

-Mình làm tắt được không bạn :/?

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2022}\)

\(\Rightarrow\dfrac{bc+ca+ab}{abc}=\dfrac{1}{a+b+c}\)

\(\Rightarrow\left(bc+ca+ab\right)\left(a+b+c\right)=abc\)

\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc=abc\)

\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc=0\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Rightarrow a=-b\) hay \(b=-c\) hay \(c=-a\)

\(\Rightarrow c=2022\) hay \(a=2022\) hay \(b=2022\)

-Nếu \(a=-b\)\(\Rightarrow B=\dfrac{1}{a^{2021}}+\dfrac{1}{b^{2021}}+\dfrac{1}{c^{2021}}=\dfrac{1}{a^{2021}}-\dfrac{1}{a^{2021}}+\dfrac{1}{2022^{2021}}=\dfrac{1}{2022^{2021}}\)

-Tương tự các trường hợp còn lại.

\(a+\dfrac{1}{a}=3\)

\(\Rightarrow\left(a+\dfrac{1}{a}\right)^2=9\)

\(\Rightarrow a^2+2+\dfrac{1}{a^2}=9\)

\(\Rightarrow a^2+\dfrac{1}{a^2}=7\)

\(a+\dfrac{1}{a}=3\)

\(\Rightarrow\left(a+\dfrac{1}{a}\right)^3=27\)

\(\Rightarrow a^3+\dfrac{1}{a^3}+3a.\dfrac{1}{a}\left(a+\dfrac{1}{a}\right)=27\)

\(\Rightarrow a^3+\dfrac{1}{a^3}+3.3=27\)

\(\Rightarrow a^3+\dfrac{1}{a^3}=18\)

\(\left(a^2+\dfrac{1}{a^2}\right)\left(a^3+\dfrac{1}{a^3}\right)=7.18=126\)

\(\Rightarrow a^5+\dfrac{1}{a}+a+\dfrac{1}{a^5}=126\)

\(\Rightarrow a^5+3+\dfrac{1}{a^5}=126\)

\(\Rightarrow a^5+\dfrac{1}{a^5}=123\)

Lời giải:

$y^2+2xy-3x-2=0$

$\Leftrightarrow y^2+2xy+x^2=x^2+3x+2$
$\Leftrightarrow (x+y)^2=(x+1)(x+2)$

Dễ thấy với mọi $x\in\mathbb{Z}$ thì $(x+1, x+2)=1$ nên để tích của chúng là scp thì $x+1, x+2$ cũng là scp

Đặt $x+1=a^2; x+2=b^2$ với $a,b\in\mathbb{Z}$
$\Rightarrow 1=b^2-a^2=(b-a)(b+a)$

$\Rightarrow b-a=b+a=1$ hoặc $b-a=b+a=-1$

$\Rightarrow a=0\Rightarrow x=-1$

Khi đó:

$(x+y)^2=(x+1)(x+2)=0$

$\Rightarrow y=-x=1$

Vậy $(x,y)=(-1,1)$

lỗi  r bn

hết lỗi chưa bạn