Ôn tập chương III - Hỏi đáp

=> \(\left[\frac{x-1}{2}+1\right]^2=1600\)

=> \(\left[\frac{x-1}{2}+1\right]^2=40^2\)

=> \(\left[\frac{x-1}{2}+1\right]=40^{ }\)

=> \(\frac{x-1}{2}=39\)

=> x - 1 = 39.2

=> x - 1 = 78

=> x=79

TB mỗi bn đc số tiền là:

[(102000 + 231000 + 177000) : 2] : 3 =85000 đ

Đ/S: 85000đ

Tổng số tiền 3 bạn là: (102000 + 231000 + 177000) :2 = 255000 đồng

Trung bình mỗi người có: 255000 : 3 = 85000 đồng

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+.....+\frac{1}{97.100}=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.......+\frac{1}{97}-\frac{1}{100}\right)=\frac{1}{3}\left(1-\frac{1}{100}\right)=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)

Gọi dãy phân số trên là A

A = \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)

A = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)

A = \(1-\frac{1}{100}\)

A = \(\frac{99}{100}\)

Đặt tử số là A

Ta có: 2A = 2 + 2² + 2³ + 2⁴+...+2²⁰⁰⁹

=> A = 2A - A= 2²⁰⁰⁹ - 1

=> B = \(\frac{2^{2009}-1}{1-2^{2009}}\) = -1

Vì 10% giá cuốn sách đó tương ứng với 1200đ nên ta có giá cuốn sách là:

1200 : 10% = 1200 : 10/100 = 12 000đ

Vậy Oanh đã mua sách với giá:

12 000 - 1200 = 10 800đ

Vì 10% giá cuốn sách đó tương ứng với 1200đ nên ta có giá cuốn sách là:

1200 : 10% = 1200 : 10/100 = 12 000đ

Vậy Oanh đã mua sách với giá:

12 000 - 1200 = 10 800đ

|||||x????
a,25%=75

⇔ x= \(75:\frac{1}{4}=75.4=300\)

\(b,\frac{11}{12}x+\frac{3}{4}=\frac{1}{6}\)

⇔ \(\frac{11}{x}x=\frac{2}{12}-\frac{9}{12}\)

⇔ \(\frac{11}{12}x=-\frac{7}{12}\)

⇔ \(x=-\frac{7}{11}\)

Vậy...

c, \(\frac{3}{4}+\frac{1}{4}\left(x-1\right)=\frac{1}{2}\)

⇔ \(\frac{1}{4}\left(x-1\right)=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\)

⇔ x - 1 = - 1

⇔ x = 0

Vậy....

\(d,\frac{1}{2}.\left|x-\frac{3}{5}\right|-\frac{1}{5}=0\)

⇔ \(\frac{1}{2}\left|x-\frac{3}{5}\right|=\frac{1}{5}\)

⇔ \(\left|x-\frac{3}{5}\right|=\frac{2}{5}\)

⇔ \(\left[{}\begin{matrix}x=\frac{2}{5}+\frac{3}{5}=1\\x=-\frac{2}{5}+\frac{3}{5}=\frac{1}{5}\end{matrix}\right.\)

Vậy....

Ta có: \(\frac{1}{2}\left(3x-\frac{1}{2}\right)-\frac{1}{4}x=7\)

\(\Leftrightarrow\frac{3}{2}x-\frac{1}{4}-\frac{1}{4}x=7\)

\(\Leftrightarrow\frac{5}{4}x-\frac{1}{4}=7\)

\(\Leftrightarrow\frac{5}{4}x=7+\frac{1}{4}=\frac{29}{4}\)

hay \(x=\frac{29}{4}:\frac{5}{4}=\frac{29}{4}\cdot\frac{4}{5}=\frac{29}{5}\)

Vậy: \(x=\frac{29}{5}\)

a, \(\left(3-7x\right)^2=\frac{1}{4}\)

⇔ \(\left[{}\begin{matrix}3-7x=\frac{1}{4}\\3-7x=-\frac{1}{4}\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}7x=\frac{11}{4}\\7x=\frac{13}{4}\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=\frac{11}{28}\\x=\frac{13}{28}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{11}{28};\frac{13}{28}\right\}\)

b, \(\left(x-1\right)^2+\left(2x-1\right)^2=0\)

Ta có \(\left(x-1\right)^2\ge0;\left(2x-1\right)^2\ge0\)\(\forall\)x

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

⇒ x ∈ ∅

c, \(5^x+5^x+2=650\)

⇒ \(2.5^x=648\)

⇒\(5^x=324\)

( có j đó sai sai :v x là số tự nhiên??)

d, \(\left(5x-1\right)\left(2x-\frac{1}{5}\right)=0\)

⇒ \(\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{4}=0\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{8}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{5};\frac{1}{8}\right\}\)

e, \(1,25-\left|2,5-x\right|=1,75\)

⇒ | 2,5 - x| = - 0,5

Ta có : | 2,5 - x| ≥ 0 \(\forall\)x

⇒ | 2,5 - x| = - 0,5 (vô lí)

⇒ x ∈ ∅

P/s: ko chắc :))

b,(x-1)^2+(2x-1)^2=0
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0.\left(2x-1\right)^2\\\left(2x-1\right)^2=0.\left(x-1\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0^2\\\left(2x-1\right)^2=0^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)=0\\\left(2x-1\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0+1\\2x=0+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1:2=\frac{1}{2}\end{matrix}\right.\)
Vậy x\(\in\left\{1;\frac{1}{2}\right\}\)

a. \(\frac{x}{3}-\frac{1}{2}=\frac{4}{9}\)

\(\frac{x}{3}=\frac{4}{9}+\frac{1}{2}\)

\(\frac{x}{3}=\frac{8}{18}+\frac{9}{18}\)

\(\frac{x}{3}=\frac{17}{18}\)

\(\Rightarrow18x=17.3=51\)

\(\Rightarrow x=51:18=\frac{17}{6}\)

Vậy \(x=\frac{17}{6}.\)

b. \(\frac{1}{5}+x:\frac{1}{2}=\frac{3}{4}\)

\(x:\frac{1}{2}=\frac{3}{4}-\frac{1}{5}\)

\(x:\frac{1}{2}=\frac{15}{20}-\frac{4}{20}\)

\(x:\frac{1}{2}=\frac{11}{20}\)

\(x=\frac{11}{20}.\frac{1}{2}\)

\(x=\frac{11}{40}\)

Vậy \(x=\frac{11}{40}.\)

Có \(\dfrac{x}{3}=\dfrac{7}{8}\)

\(\Rightarrow8x=7.3\)

\(\Rightarrow8x=21\)

\(\Rightarrow x=\frac{21}{8}\)

\(\frac{x}{3}=\frac{7}{8}\)

=> 8x=7.3

=> 8x=21

=> x=\(\frac{21}{8}\)

S=1/1.3+1/3.5+1/5.7+....+1/2017.2018

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

=\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2019}\right)\)

=\(\frac{1}{2}-\frac{2018}{2019}\)

=\(\frac{-2017}{4038}\)

Lời giải:

$\frac{1}{x}+\frac{1}{y}=\frac{1}{6}$

$\Rightarrow 6(x+y)=xy$

$\Leftrightarrow 6x+6y-xy=0$

$\Leftrightarrow x(6-y)-6(6-y)=-36$

$\Leftrightarrow (x-6)(6-y)=-36$

Vì $x,y$ nguyên dương nên $x-6>-6; 6-y< 6$
Xét các TH sau:

$x-6=1; 6-y=-36\Rightarrow x=7; y=42$

$x-6=2; 6-y=-18\Rightarrow x=8; y=24$

$x-6=3; 6-y=-12\Rightarrow x=9; y=18$

$x-6=4; 6-y=-9\Rightarrow x=10; y=15$

...... còn các TH còn lại xét tương tự.