Ôn tập chương 1

Ta có:\(1-\dfrac{a}{b}=\dfrac{b-a}{b}\)

\(1-\dfrac{a+1}{b+1}=\dfrac{b+1-a-1}{b+1}=\dfrac{b-a}{b+1}\)

*)Nếu b>a=>b-a>0

=>\(\dfrac{b-a}{b}>\dfrac{b-a}{b+1}\)

=>\(\dfrac{a}{b}< \dfrac{a+1}{b+1}\)

*)Nếu b=a=>\(\dfrac{a}{b}=\dfrac{a+1}{b+1}=1\)

*)Nếu b<a =>b-a<0

=>\(\dfrac{b-a}{a}< \dfrac{b-a}{b+1}\)

=>\(\dfrac{a}{b}>\dfrac{a+1}{b+1}\)

Vậy...

Bài 1:

Với mọi số hữu tỉ ta luôn có: \(\left\{{}\begin{matrix}x\le\left|x\right|\\-x\le\left|x\right|\end{matrix}\right.\) và \(\left\{{}\begin{matrix}y\le\left|y\right|\\-y\le\left|y\right|\end{matrix}\right.\)

Cộng từng đẳng thức lại \(\Rightarrow\left\{{}\begin{matrix}x+y\le\left|x\right|+\left|y\right|\\-x-y\le\left|x\right|+\left|y\right|\end{matrix}\right.\)

Hay: \(\left\{{}\begin{matrix}x+y\le\left|x\right|+\left|y\right|\\x+y\ge-\left(\left|x\right|+\left|y\right|\right)\end{matrix}\right.\)\(\Leftrightarrow-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)

Vậy \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)

Dấu bằng xảy ra khi \(xy=0\)

Câu b tương tự nhé.

Bài 2:

Ta có:

\(A=\left|x-2001\right|+\left|x-1\right|=\left|2001-x\right|+\left|1-x\right|\ge\left|2001-x+x-1\right|=2000\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}2001-x\ge0\\x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow2001\ge x\ge1\)

Vậy \(_{min}A=2000\) khi \(2001\ge x\ge1\)

Bài 2:

Ta có: \(A=\left|x-2001\right|+\left|x-1\right|=\left|2001-x\right|+\left|x-1\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:

\(A\ge\left|2001-x+x-1\right|=\left|2000\right|=2000\)

Dấu " = " khi \(\left\{{}\begin{matrix}2001-x\ge0\\x-1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le2001\\x\ge1\end{matrix}\right.\)

Vậy \(MIN_A=2000\) khi \(1\le x\le2001\)

Bài 2:

\(A=\left|x-2001\right|+\left|x-1\right|=\left|2001-x\right|+\left|x-1\right|\)

(do \(\left|A\left(x\right)\right|=\left|-A\left(x\right)\right|\))

Với mọi giá trị của \(x\in R\) ta có:

\(\left|2001-x\right|\ge2001-x;\left|x-1\right|\ge x-1\)

\(\Rightarrow\left|2001-x\right|+\left|x-1\right|\ge2001-x+x-1\)

\(\Rightarrow\left|2001-x\right|+\left|x-1\right|\ge2000\)

Hay \(A\ge2000\) với mọi giá trị của \(x\in R\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2001-x\ge0\\x-1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le2001\\x\ge1\end{matrix}\right.\)

\(\Rightarrow1\le x\le2001\)

Vậy GTNN của biểu thức A là 2000 đạt được khi và chỉ khi \(1\le x\le2001\)

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

Với \(x\ge\dfrac{1}{2}\) thì \(x-\dfrac{1}{2}\ge0\) nên \(\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2},\) thay vào A ta có:

\(A=x-\dfrac{1}{2}+\dfrac{3}{4}-x=\dfrac{1}{4}\)

Với \(x< \dfrac{1}{2}\) thì \(x-\dfrac{1}{2}< 0\) nen \(\left|x-\dfrac{1}{2}\right|=-x+\dfrac{1}{2},\) thay vào A ta có:

\(A=-x+\dfrac{1}{2}+\dfrac{3}{4}-x=-2x+\dfrac{5}{4}\)

b) Với \(x\ge\dfrac{1}{2}\) thì \(A=\dfrac{1}{4}\) _______( 1 )_______

Với \(x< \dfrac{1}{2}\) thì \(-2x>-1\Leftrightarrow-2x+\dfrac{5}{4}>-1+\dfrac{5}{4}\) hay \(A>\dfrac{1}{4}\) __________( 2 )_________

Từ ( 1 ) và ( 2 ) suy ra \(A\ge\dfrac{1}{4}\)

Vậy \(GTNN\left(A\right)=\dfrac{1}{4}\) khi \(x\ge\dfrac{1}{2}\)

Giải:

a) \(\left|x\right|=2,5\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2,5\\x=-2,5\end{matrix}\right.\)

b) \(\left|x\right|=-1,2\)

Vì \(\left|x\right|\ge0\forall x\)

\(\Leftrightarrow x\in\varnothing\)

c) \(\left|x\right|+0,573=2\)

\(\Leftrightarrow\left|x\right|=2-0,573=1,427\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1,427\\x=-1,427\end{matrix}\right.\)

d) \(\left|x+\dfrac{1}{3}\right|-4=-1\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-\dfrac{1}{3}\\x=-3-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2,\left(6\right)\\x=-3,\left(3\right)\end{matrix}\right.\)

e) \(1,5\left|3x-1\right|+4,659=9,103\)

\(\Leftrightarrow1,5\left|3x-1\right|=9,103-4,659=4,444\)

\(\Leftrightarrow\left|3x-1\right|=\dfrac{4,444}{1,5}=2,962\left(6\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-1=2,962\left(6\right)\\3x-1=-2,962\left(6\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=2,962\left(6\right)-1\\3x=-2,962\left(6\right)-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=1,962\left(6\right)\\3x=-3,962\left(6\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1,962\left(6\right)}{3}\\x=\dfrac{-3,962\left(6\right)}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,654\left(2\right)\\x=-1,320\left(8\right)\end{matrix}\right.\)

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P/s: Chú ý câu cuối có nhiều dữ kiện phức tạo nên không thể chắc chắc đúng 100%

Câu a,b,c,d link đây Câu hỏi của (っ◔◡◔)っ ♥ GDragon Huyền Tồ ♥

e) \(\left|3x-1\right|=\left(9,103-4,659\right):\dfrac{3}{2}\)

\(\left|3x-1\right|=4,444:\dfrac{3}{2}\)

\(\left|3x-1\right|=\dfrac{4444}{1000}:\dfrac{3}{2}\)

\(\left|3x-1\right|=\dfrac{4444.2}{1000.3}=\dfrac{4444}{1500}\)

____* \(3x-1=\dfrac{4444}{1500}\)

\(3x=\dfrac{4444}{1500}+1\)

\(3x=\dfrac{5944}{1500}\)

\(x=\dfrac{5944}{4500}=\dfrac{1486}{1125}\)

____* \(3x-1=-\dfrac{4444}{1500}\)

\(3x=-\dfrac{4444}{1500}+1\)

\(x=-\dfrac{2944}{4500}=-\dfrac{736}{1125}\)

P/s: ( Nếu có sai chỗ nào thì sửa giùm nha đang cấn trận đánh bang bang hay rồi nên phải nhanh )

đễ

đề yêu câu ftinhs hay tính nhanh Đoàn Đức Hiếu cho nhok cày GP nè

a) \(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}^2\right).\left(-1\right)^5\)

\(=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)\)

\(=\dfrac{1}{6}.\left(-1\right)\)

=\(\dfrac{-1}{6}\)

b) \(\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2\)

\(=\dfrac{4}{25}.\dfrac{25}{144}\)

\(=\dfrac{1}{36}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow a=bk;c=dk\)

\(\dfrac{a+b}{a}=\dfrac{bk+b}{bk}=\dfrac{b\left(k+1\right)}{bk}=\dfrac{k+1}{k}\)

\(\dfrac{c+d}{c}=\dfrac{dk+d}{dk}=\dfrac{d\left(k+1\right)}{dk}=\dfrac{k+1}{k}\)

\(\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\rightarrowđpcm\)

\(\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{bk}=\dfrac{k-1}{k}\)

\(\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\)

\(\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\rightarrowđpcm\)

\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\rightarrowđpcm\)

\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\)

\(\Rightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\rightarrowđpcm\)

\(\dfrac{72-x}{7}=\dfrac{x-4}{9}\)

\(\Rightarrow9\left(72-x\right)=7\left(x-4\right)\)

\(\Rightarrow648-9x=2x-28\)

\(\Rightarrow11x-28=648\)

\(\Rightarrow11x=676\Rightarrow x=\dfrac{676}{11}\)

\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)

\(\Rightarrow7\left(37-x\right)=3\left(x+13\right)\)

\(\Rightarrow259-7x=3x+39\)

\(\Rightarrow10x+39=259\)

\(\Rightarrow10x=220\Rightarrow x=22\)

\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

\(\Rightarrow\left(x+4\right)^2=100\)

\(\Rightarrow\left(x+4\right)^2=\pm10^2\)

\(\Rightarrow\left[{}\begin{matrix}x+4=10\Rightarrow x=6\\x+4=-10\Rightarrow x=-14\end{matrix}\right.\)

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

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

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

\(\Rightarrow x^2+3x-x-3=x^2+2x-2x-4\)

\(\Rightarrow x^2+2x-3=x^2-4\)

\(\Rightarrow2x-3=-4\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\dfrac{1}{2}\)

(-0,246).\(3\dfrac{2}{7}\).(-3)\(^2\)

= (-0,246).\(\dfrac{23}{7}\).9

=\(\dfrac{-2829}{3500}\).9

=\(\dfrac{-25461}{3500}\)

( - 0,246) . \(3\dfrac{2}{7}\). (-3)²

^{= ( - 0,246) . \(\dfrac{23}{7}\) . 9}

^{= \(\dfrac{-2829}{3500}\)} . 9

= \(-\dfrac{25461}{3500}\)

Chúc bạn học tốt

P/S: Bài dễ thế mà còn đăng lên hả trời

pn ơi hình như đề sai a+5/a-5 va b+6/b-6

ta có : a+5/a-5=b+6/b-6
=> a+5/b+6=a-5/b-6
áp dụng dãy tỉ số bằng nhau ta được:
a+5/b+6=a-5/b-6 =(a+5+a-5)/(b+6+b-6)=(a+5-a+5)/(b+6-b+6)
=> 2a/2b = 10/12
=> a/b = 5/6