b: Bảng giá trị:

Vẽ đồ thị:

(123/233+1995/2015+121212/242424)*(1/4-1/5-1/20)

=(123/233+399/403+1/2)*(5/20-4/20-1/20)

=(123/233+399/403+1/2)*0

=0

=8364861346377613674163463961267236319639692649461292639463296366313713676767676677676367363636343634673463367436746767343873467647

Ta có :\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}=2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)

\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{15\times16}\right)\)

\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)=2\times\left(\frac{1}{2}-\frac{1}{16}\right)=\frac{7}{8}\)

c.

\(\left\{\begin{matrix} 9x-6y=4\\ 15x-10y=7\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{6y+4}{9}\\ 15x-10y=7\end{matrix}\right.\)

\(\Rightarrow 15.\frac{6y+4}{9}-10y=7\)

\(\Leftrightarrow \frac{5}{3}(6y+4)-10y=7\Leftrightarrow \frac{20}{3}=7\) (vô lý)

Do đó hpt vô nghiệm.

d.

\(\left\{\begin{matrix} 4x+5y=3\\ x-3y=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 4x+5y=3\\ x=3y+5\end{matrix}\right.\Rightarrow 4(3y+5)+5y=3\)

\(\Leftrightarrow 17y+20=3\Leftrightarrow 17y=-17\Leftrightarrow y=-1\)

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

Vậy HPT có nghiệm $(x,y)=(2,-1)$

Các câu còn lại bạn làm theo pp tương tự.

1.

HPT \(\Leftrightarrow \left\{\begin{matrix} 5x-y=4\\ 3x-y=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 5x-y=4\\ y=3x-5\end{matrix}\right.\)

\(\Rightarrow 5x-(3x-5)=4\Leftrightarrow 2x+5=4\Leftrightarrow 2x=-1\Leftrightarrow x=\frac{-1}{2}\)

\(y=3x-5=\frac{-3}{2}-5=\frac{-13}{2}\)

Vậy HPT có nghiệm $(x,y)=(\frac{-1}{2}, \frac{-13}{2})$

\(\dfrac{3}{5}\) x \(\dfrac{1}{7}\) x 1\(\dfrac{1}{2}\)

= \(\dfrac{3}{35}\) x \(\dfrac{3}{2}\)

= \(\dfrac{9}{70}\)

Bài 2:

a: Xét ΔBAD và ΔBFD có

BA=BF

\(\hat{ABD}=\hat{FBD}\)

BD chung

Do đó: ΔBAD=ΔBFD

=>DA=DF

b: ΔBAD=ΔBFD

=>\(\hat{BAD}=\hat{BFD}\)

mà \(\hat{BAD}+\hat{DAE}=180^0\) (hai góc kề bù)

và \(\hat{BFD}+\hat{DFC}=180^0\) (hai góc kề bù)

nên \(\hat{DAE}=\hat{DFC}\)

Ta có: BA+AE=BE

BF+FC=BC

mà BA=BF và BE=BC

nên AE=CF

Xét ΔDAE và ΔDFC có

AE=FC

\(\hat{DAE}=\hat{DFC}\)

DA=DF

Do đó: ΔDAE=ΔDFC

c: ΔDAE=ΔDFC

=>\(\hat{ADE}=\hat{FDC}\)

mà \(\hat{FDC}+\hat{FDA}=180^0\) (hai góc kề bù)

nên \(\hat{ADE}+\hat{FDA}=180^0\)

=>E,D,F thẳng hàng

g: \(x-\frac{7}{12}x+\frac38x=\frac{5}{24}\)

=>\(x\left(1-\frac{7}{12}+\frac38\right)=\frac{5}{24}\)

=>\(x\left(\frac{24}{24}-\frac{14}{24}+\frac{9}{24}\right)=\frac{5}{24}\)

=>\(x\cdot\frac{19}{24}=\frac{5}{24}\)

=>\(x=\frac{5}{24}:\frac{19}{24}=\frac{5}{19}\)

h: \(\left(x-\frac13\right)^2-\frac12=1\frac34\)

=>\(\left(x-\frac13\right)^2-\frac12=\frac74\)

=>\(\left(x-\frac13\right)^2=\frac12+\frac74=\frac94\)

=>\(\left[\begin{array}{l}x-\frac13=\frac32\\ x-\frac13=-\frac32\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32+\frac13=\frac{9+2}{6}=\frac{11}{6}\\ x=\frac{-3}{2}+\frac13=\frac{-9+2}{6}=-\frac76\end{array}\right.\)

3n + 1⋮ 2n + 1 đk n \(\ne\) - \(\dfrac{1}{2}\)

6n + 2 ⋮ 2n + 1

3.(2n + 1) - 1 ⋮ 2n + 1

                   1 ⋮ 2n + 1

2n + 1 \(\in\) { -1; 1}

n \(\in\) {-1; 0}

\(\dfrac{3}{5}\times\dfrac{4}{7}\times1\dfrac{1}{2}\)

\(=\dfrac{3}{5}\times\dfrac{4}{7}\times\dfrac{3}{2}\)

\(=\dfrac{3\times4\times3}{5\times7\times2}\)

\(=\dfrac{3\times2\times3}{5\times7}\)

\(=\dfrac{18}{35}\)