Bài 2: 

\(a,x-\dfrac{3}{10}=\dfrac{7}{15}\cdot\dfrac{3}{5}\\ =>x-\dfrac{3}{10}=\dfrac{7}{25}\\ =>x=\dfrac{7}{25}+\dfrac{3}{10}\\ =>x=\dfrac{29}{50}\\ b.2x+\dfrac{3}{2}=\dfrac{-2}{5}\\ =>2x=\dfrac{-2}{5}-\dfrac{3}{2}\\ =>x=\dfrac{-4}{10}-\dfrac{15}{10}=\dfrac{-19}{10}\\ =>x=\dfrac{-19}{10}:2=-\dfrac{19}{20}\\ c,\left(x-\dfrac{1}{2}\right)^3=-8\\ =>\left(x-\dfrac{1}{2}\right)^3=\left(-2\right)^3\\ =>x-\dfrac{1}{2}=-2\\ =>x=-2+\dfrac{1}{2}\\ =>x=\dfrac{-3}{2}\\ d,\left(\dfrac{7}{5}\right)^x=\dfrac{49}{25}\\ =>\left(\dfrac{7}{5}\right)^x=\left(\dfrac{7}{5}\right)^2\\=>x=2\)

Câu 3:

a: \(\widehat{xOz}+\widehat{yOz}=180^0\)(hai góc kề bù)

=>\(\widehat{yOz}=180^0-\widehat{xOz}=180^0-60^0=120^0\)

b: Ot là phân giác của góc yOz

=>\(\widehat{yOt}=\widehat{zOt}=\dfrac{\widehat{yOz}}{2}=\dfrac{120^0}{2}=60^0\)

Ta có: \(\widehat{xOt}+\widehat{yOt}=180^0\)(hai góc kề bù)

=>\(\widehat{xOt}+60^0=180^0\)

=>\(\widehat{xOt}=120^0\)

c: Ta có: \(\widehat{xOm}=\widehat{yOt}\)(hai góc đối đỉnh)

mà \(\widehat{yOt}=60^0\)

nên \(\widehat{xOm}=60^0\)

Ta có: \(\widehat{xOm}=\widehat{xOz}\left(=60^0\right)\)

=>Ox là phân giác của góc mOz

Câu 1:

b: \(\dfrac{11}{2}\cdot4\dfrac{5}{3}-2\dfrac{5}{3}\cdot\dfrac{11}{2}\)

\(=\dfrac{11}{2}\left(4+\dfrac{5}{3}-2-\dfrac{5}{3}\right)\)

\(=\dfrac{11}{2}\cdot2=11\)

d: \(\left(\dfrac{3}{7}\right)^0\cdot1^{15}+\dfrac{7}{9}:\left(\dfrac{2}{3}\right)^2-\dfrac{4}{5}\)

\(=1\cdot1+\dfrac{7}{9}:\dfrac{4}{9}-\dfrac{4}{5}\)

\(=1-\dfrac{4}{5}+\dfrac{7}{4}=\dfrac{1}{5}+\dfrac{7}{4}=\dfrac{4}{20}+\dfrac{35}{20}=\dfrac{39}{20}\)

Bài 6:

a) Ta có:

\(\sqrt{x+7}\ge0\forall\left(x\ge-7\right)\\ =>2\sqrt{x+7}\ge0\forall\left(x\ge-7\right)\\ =>A=2\sqrt{x+7}-5\ge-5\forall\left(x\ge-7\right)\)

Dấu "=" xảy ra: `x+7=0`

`<=>x=-7` 

b) Ta có:

\(\sqrt{x-8}\ge0\forall\left(x\ge8\right)\\ =>\dfrac{1}{2}\sqrt{x-8}\ge0\forall\left(x\ge8\right)\\ =>A=-12+\dfrac{1}{2}\sqrt{x-8}\ge0\forall\left(x\ge8\right)\)

Dấu "=" xảy ra: `x-8<=>x=8` 

Bài 7:

a: ĐKXĐ: x>=0

\(-\dfrac{1}{4}\sqrt{x}< =0\forall x\) thỏa mãn ĐKXĐ

=>\(B=-\dfrac{1}{4}\sqrt{x}+4< =4\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

b: ĐKXĐ: \(\left[{}\begin{matrix}x>=2\\x< =-2\end{matrix}\right.\)

\(\sqrt{x^2-4}>=0\forall x\) thỏa mãn DKXĐ
=>\(-\dfrac{1}{4}\sqrt{x^2-4}< =0\forall x\) thỏa mãn ĐKXĐ

=>\(B=-\dfrac{1}{4}\sqrt{x^2-4}+1< =1\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(x^2-4=0\)

=>\(x^2=4\)

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

a: 25-(5-x)=-7

=>5-x=25+7=32

=>x=5-32=-27

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

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

=>\(x-1=\dfrac{2}{3}:\dfrac{1}{4}=\dfrac{2}{3}\cdot4=\dfrac{8}{3}\)

=>\(x=\dfrac{8}{3}+1=\dfrac{11}{3}\)

d: \(\left(2x+1\right)\left(x-\dfrac{1}{7}\right)=0\)

=>\(\left[{}\begin{matrix}2x+1=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{7}\end{matrix}\right.\)

c: \(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)

=>\(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}=\dfrac{7}{21}-\dfrac{9}{21}=-\dfrac{2}{21}\)

=>\(x=-\dfrac{2}{21}:\dfrac{4}{7}=-\dfrac{2}{21}\cdot\dfrac{7}{4}=\dfrac{-1}{6}\)

a: Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

mà \(\dfrac{\widehat{A}}{1}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{5}\)

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{\widehat{A}}{1}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{5}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{1+3+5}=\dfrac{180^0}{9}=20^0\)

=>\(\widehat{A}=20^0;\widehat{B}=60^0;\widehat{C}=100^0\)

b: BD là phân giác góc ngoài tại B

=>\(\widehat{CBD}=\dfrac{180^0-60^0}{2}=\dfrac{120^0}{2}=60^0\)

\(\widehat{BCD}+\widehat{BCA}=180^0\)

=>\(\widehat{BCD}+100^0=180^0\)

=>\(\widehat{BCD}=80^0\)

Xét ΔCBD có \(\widehat{CBD}+\widehat{BCD}+\widehat{BDC}=180^0\)

=>\(\widehat{ADB}+80^0+60^0=180^0\)

=>\(\widehat{ADB}=40^0\)

Thay a=1/3;b=3/5 vào A, ta được:

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

\(=1-\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{4}{5}+\dfrac{1}{10}=\dfrac{9}{10}\)

\(C=\dfrac{5}{2\cdot7}+\dfrac{16}{7\cdot9}-\dfrac{2}{9\cdot11}-\dfrac{29}{1\cdot11}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{29}{11}\)

\(=\dfrac{1}{2}-\dfrac{28}{11}=\dfrac{11-56}{22}=\dfrac{-45}{22}< \dfrac{1}{3}\)

67

Ta có:

\(\left|x-5\right|+\left|2-4x\right|=\left|x-5\right|+\left|\dfrac{1}{2}-x\right|+3\left|\dfrac{1}{2}-x\right|\)

Mà \(\left|x-5\right|+\left|\dfrac{1}{2}-x\right|\ge\left|x-5+\dfrac{1}{2}-x\right|=\dfrac{9}{2}>4\)

\(\Rightarrow\left|x-5\right|+\left|2-4x\right|\ge4+3\left|\dfrac{1}{2}-x\right|>4>3\)

Vậy pt đã cho vô nghiệm

a: Xét ΔAMB và ΔAMC có

AM chung

MB=MC

AB=AC

Do đó: ΔAMB=ΔAMC

b: ΔAMB=ΔAMC

=>\(\widehat{AMB}=\widehat{AMC}\)

mà \(\widehat{AMB}+\widehat{AMC}=180^0\)(hai góc kề bù)

nên \(\widehat{AMB}=\widehat{AMC}=\dfrac{180^0}{2}=90^0\)

=>AD\(\perp\)BC tại M

Xét ΔMAB vuông tại M và ΔMDC vuông tại M có

MA=MD

MB=MC

Do đó: ΔMAB=ΔMDC
=>\(\widehat{MAB}=\widehat{MDC}\)

=>AB//DC

c: ta có: ME\(\perp\)AB

AB//CD

Do đó: ME\(\perp\)CD

mà MF\(\perp\)CD

và ME,MF có điểm chung là M
nên M,E,F thẳng hàng

Xét ΔMEB vuông tại E và ΔMFC vuông tại F có

MB=MC

\(\widehat{MBE}=\widehat{MCF}\)(cmt)

Do đó: ΔMEB=ΔMFC

=>ME=MF

=>M là trung điểm của EF

TH1: \(-\dfrac{1}{2}\le x\le\dfrac{5}{3}\)

\(\left(2x+1\right)+\left(5-3x\right)=6\\ =>2x+1+5-3x=6\\ =>\left(2x-3x\right)+6=6\\ =>x=0\left(tm\right)\)

TH2: \(x>\dfrac{5}{3}\)

\(\left(2x+1\right)-\left(5-3x\right)=6\\ =>2x+1-5+3x=6\\ =>2x+3x=6-1+5\\ =>5x=10\\ =>x=\dfrac{10}{5}=2\left(tm\right)\)

TH3: \(x< -\dfrac{1}{2}\)

\(-\left(2x+1\right)+\left(5-3x\right)=6\\ =>-2x-1+5-3x=6\\ =>-2x-3x+4=6\\ =>-5x=6-4=2\\ =>x=-\dfrac{2}{5}\left(ktm\right)\)

Ta có: \(\widehat{xMy'}=\widehat{x'My}\)(hai góc đối đỉnh)

mà \(\widehat{xMy'}=90^0\)

nên \(\widehat{x'My}=90^0\)

Ta có: \(\widehat{xMy'}+\widehat{x'My'}=180^0\)(hai góc kề bù)

=>\(\widehat{x'My'}=180^0-90^0=90^0\)

Ta có: \(\widehat{xMy}=\widehat{x'My'}\)(hai góc đối đỉnh)

mà \(\widehat{x'My'}=90^0\)

nên \(\widehat{xMy}=90^0\)