Ôn tập: Phương trình bâc nhất một ẩn

1a)\(\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)

=> \(\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)

=>\(\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)

=> x=100( vi \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\ne0\)

b) \(\dfrac{x-5}{2012}-1+\dfrac{x-4}{2013}-1=\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1\)

=> \(\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}-\dfrac{x-2017}{2015}=0\)

=>(x-2017).\(\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)

=> x=2015(vi .....................................................≠0)

2)

3) A=(a⁴-2a³+a²)+(2a²-4a+2)+3=(a²-a)²+2(a-1)²+3≥3

Dau bang xay ra khi : \(\left\{{}\begin{matrix}a^2-a=0\\a-1=0\end{matrix}\right.\)

=> a=1

Vay GTNN cua A=3 khi a=1

\(\Leftrightarrow\dfrac{x+9}{10}+1+\dfrac{x+10}{9}+1=\dfrac{9}{x+10}+1+\dfrac{10}{x+9}+1\)

\(\Leftrightarrow\dfrac{x+19}{10}+\dfrac{x+19}{9}=\dfrac{x+19}{x+10}+\dfrac{x+19}{x+9}\)

\(\Leftrightarrow\left(x+9\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{x+10}-\dfrac{1}{x+9}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+9=0\\\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{x+10}-\dfrac{1}{x+9}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-9\\x=0\end{matrix}\right.\)

Bạn nhớ ĐKXĐ

a) [x(x+1].[(x-1)(x+2)]=24

(x²+x)(x²+x+2)=24

Dat x²+x=a , ta dc: a(a+2)=24

=> a²+2a-24=0

=> (a-4)(a+6)=0

=> a=4 hoac a=-6

Thay vao roi tu tim x nha

b)

1, \(\left(x-1\right)^2=4\)

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

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

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

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

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

\(\Rightarrow S=\left\{3;-1\right\}\)

2, \(\left(x-1\right)^2=9\left(x+1\right)^2\)

\(\Leftrightarrow\left(x-1\right)^2-9\left(x+1\right)^2=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=4\\4x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{2}\end{matrix}\right.\)

\(\Rightarrow S=\left\{-2;\dfrac{-1}{2}\right\}\)

3, \(x\left(x-1\right)=-x\left(x+3\right)\)

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

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

\(\Leftrightarrow2x^2+2x=0\)

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

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

\(\Rightarrow S=\left\{0;-1\right\}\)

4, \(\left(2x-1\right)^2=9\)

\(\Leftrightarrow\left(2x-1\right)^2-9=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Xin lỗi nhiều nha.

Máy tính mik bị sao í

2. \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow\)\(x^2+9x+x+9=x^2+5x+3x+15\)

\(\Leftrightarrow x^2+9x+x-x^2-5x-3x=15-9\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=\dfrac{6}{2}\Rightarrow x=3\)

\(S=\left\{3\right\}\)

\(1,5-\left(6-x\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-6+x=12-8x\)

\(\Leftrightarrow x+8x=12-5+6\)

\(\Leftrightarrow9x=13\)

\(\Leftrightarrow x=\dfrac{13}{9}\)

Vậy tập nghiệm của pt là \(S=\left\{\dfrac{13}{9}\right\}\)

\(2,\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow x^2+10x+9=x^2+8x+15\)

\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)

\(\Leftrightarrow2x-6=0\)

\(\Leftrightarrow x=3\)

Vậy tập nghiệm của pt là S = { 3 }

\(3,\dfrac{3\left(5x-2\right)}{4}-2=\dfrac{7x}{3}-5\left(x-7\right)\)

\(\Leftrightarrow\dfrac{9\left(5x-2\right)-24}{12}=\dfrac{28x-60\left(x-7\right)}{12}\)

\(\Rightarrow45x-18-24=28x-60x+420\)

\(\Leftrightarrow45x-28x+60x=420+18+24\)

\(\Leftrightarrow77x=462\)

\(\Leftrightarrow x=6\)

Vậy tập nghiệm của pt là S = { 6 }

\(4,3\left(x+1\right)\left(2x+5\right)=3\left(x+1\right)\left(7x-4\right)\)

\(\Leftrightarrow3\left(x+1\right)\left(2x+5\right)-3\left(x+1\right)\left(7x-4\right)=0\)

\(\Leftrightarrow3\left(x+1\right)\left(2x+5-7x+4\right)=0\)

\(\Leftrightarrow3\left(x+1\right)\left(-5x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\-5x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{9}{5}\end{matrix}\right.\)

Vậy tập nghiệm của pt là \(S=\left\{-1;\dfrac{9}{5}\right\}\)

\(5,\left(x-2\right)^2-\left(3x+1\right)^2+x\left(4x-1\right)=0\)

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

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

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

\(\Leftrightarrow\left(4x-1\right)\left(-x-3\right)=0\)

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

Vậy tập nghiệm của pt là \(S=\left\{\dfrac{1}{4};-3\right\}\)