6:

\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

mà 8<9

nên \(2^{225}< 3^{150}\)

4: \(\left|5x+3\right|>=0\forall x\)

=>\(-\left|5x+3\right|< =0\forall x\)

=>\(-\left|5x+3\right|+5< =5\forall x\)

Dấu = xảy ra khi 5x+3=0

=>x=-3/5

1:

\(\left(2x+1\right)^4>=0\)

=>\(\left(2x+1\right)^4+2>=2\)

=>\(M=\dfrac{3}{\left(2x+1\right)^4+2}< =\dfrac{3}{2}\)

Dấu = xảy ra khi 2x+1=0

=>x=-1/2

\(c,-\dfrac{8}{13}+\left(-\dfrac{7}{5}-x\right)=-\dfrac{1}{2}\\ -\dfrac{7}{5}-x=-\dfrac{1}{2}-\dfrac{8}{13}\\ -\dfrac{7}{5}-x=-\dfrac{29}{26}\\ x=-\dfrac{7}{5}-\left(-\dfrac{29}{26}\right)=-\dfrac{37}{130}\\ d,-1\dfrac{1}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{8}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{8}{7}-\left(-\dfrac{4}{21}\right)\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{20}{21}\\ x-\dfrac{7}{3}=-\dfrac{20}{21}-\left(-\dfrac{5}{3}\right)\\ x-\dfrac{7}{3}=\dfrac{5}{7}\\ x=\dfrac{5}{7}+\dfrac{7}{3}=\dfrac{64}{21}\\ e,-\dfrac{2}{3}-x:\dfrac{1}{2}=\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{2}{3}-\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{16}{15}\\ x=-\dfrac{16}{15}\times\dfrac{1}{2}=-\dfrac{8}{15}\)

c: -8/13+(-7/5-x)=-1/2

=>x+7/5+8/13=1/2

=>x=1/2-7/5-8/13=-197/130

d: \(\Leftrightarrow-\dfrac{8}{7}+\dfrac{5}{3}-\left(x-\dfrac{7}{3}\right)=\dfrac{-4}{21}\)

=>\(x-\dfrac{7}{3}=\dfrac{-8}{7}+\dfrac{5}{3}+\dfrac{4}{21}=\dfrac{-24+35+4}{21}=\dfrac{18}{21}=\dfrac{6}{7}\)

=>x=6/7+7/3=18/21+49/21=67/21

e: =>x:1/2=-2/3-2/5=-16/15

=>x=-16/15*1/2=-8/15

f: =>-8/5*x=-1/3+4/9=1/9

=>x=-1/9:8/5=-1/9*5/8=-5/72

g: =>-4/5x-1/4+x=-13/3

=>1/5x=-13/3+1/4=-52/12+3/12=-49/12

=>x=-49/12*5=-245/12

h: =>12/7:x-1/2=0 hoặc 2/5x-3/2=0

=>12/7:x=1/2 hoặc 2/5x=3/2

=>x=12/7:1/2=24/7 hoặc x=3/2:2/5=3/2*5/2=15/4

a) Ta có:

\(\widehat{yOu}+\widehat{xOy}=180^o\) (kề bù) 

\(\Rightarrow\widehat{yOu}=180^o-\widehat{xOy}\)

\(\Rightarrow\widehat{yOu}=180^o-60^o=120^o\) 

Mà: \(\widehat{xOt}+\widehat{tOu}=180^o\) (kề bù) 

\(\Rightarrow\widehat{xOt}=180^o-\widehat{tOu}\)

\(\Rightarrow\widehat{xOt}=180^o-30^o=150^o\)

b) Ta có:

\(\widehat{xOy}+\widehat{yOt}+\widehat{tOu}=\widehat{xOu}\)

\(\Rightarrow\widehat{yOt}=\widehat{xOu}-\widehat{xOy}-\widehat{tOu}\)

\(\Rightarrow\widehat{yOt}=180^o-60^o-30^o\)

\(\Rightarrow\widehat{yOt}=90^o\)

1: \(\sqrt{11}\) là số vô tỉ

2:

a: 4,9(18)=4,91818...

mà 4,91818<4,928

nên 4,9(18)<4,928

b: 4,315<4,318

=>-4,315>-4,318

=>-4,315...>-4,318...

c: \(\sqrt{3}=\sqrt{\dfrac{6}{2}}< \sqrt{\dfrac{7}{2}}\)

3: 

a: \(6=\sqrt{3};-1,7=-\sqrt{2,89}\)

0<2,89<3

=>\(0< \sqrt{2,89}< \sqrt{3}\)

=>\(-\sqrt{3}< -\sqrt{2,89}< 0\)

0<35<36<47

=>\(0< \sqrt{35}< \sqrt{36}< \sqrt{47}\)

=>\(-\sqrt{3}< -\sqrt{2,89}< 0< \sqrt{35}< \sqrt{36}< \sqrt{47}\)

=>\(-\sqrt{3}< -\sqrt{2,89}< 0< \sqrt{35}< 6< \sqrt{47}\)

b: \(-\sqrt{2\dfrac{1}{3}}=-\sqrt{2,\left(3\right)}\)

\(-1,5=-\sqrt{2,25}\)

2,25<2,3<2,(3)

=>\(\sqrt{2.25}< \sqrt{2.3}< \sqrt{2.\left(3\right)}\)

=>\(0>-1.5>-\sqrt{2.3}>-\sqrt{2\dfrac{1}{3}}\)

\(0< \sqrt{5\dfrac{1}{6}}=\sqrt{5,1\left(6\right)}< \sqrt{5,3}\)

=>\(\sqrt{5.3}>\sqrt{5\dfrac{1}{6}}>0>-1.5>-\sqrt{2.3}>-\sqrt{2\dfrac{1}{3}}\)

`#1231.2021`

`1.`

Ta có:

`y` tỉ lệ nghịch với `x` theo hệ số tỉ lệ `-4`

`=> y = (-4)/x` `(1)`

`x` tỉ lệ nghịch với `z` theo hệ số tỉ lệ `3/4`

`=> x = 3/4 \div z` `(2)`

Thay `(2)` vào `(1)`

`=> y = (-4)/(3/4 \div z) => y = -16/3 * z`

Vậy, `y` và `z` tỉ lệ thuận với nhau theo hệ số tỉ lệ `-16/3`

`=> A.`

Chọn D

\(a,x+\dfrac{1}{2}=\dfrac{3}{4}\\ x=\dfrac{3}{4}-\dfrac{1}{2}\\ x=\dfrac{1}{2}\\ b,-\dfrac{2}{3}-x=1\\x=-\dfrac{2}{3}-1\\ x=-\dfrac{5}{3}\\ d,\dfrac{1}{4}+\dfrac{3}{4}:x=\dfrac{5}{2}\\ \dfrac{3}{4}:x=\dfrac{5}{2}-\dfrac{1}{4}\\ \dfrac{3}{4}:x=\dfrac{9}{4}\\ x=\dfrac{3}{4}:\dfrac{9}{4}\\ x=\dfrac{1}{3}\\ e,\left(x+\dfrac{1}{4}\right)\cdot\dfrac{3}{4}=-\dfrac{5}{8}\\ x+\dfrac{1}{4}=-\dfrac{5}{8}:\dfrac{3}{4}\\ x+\dfrac{1}{4}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{1}{4}\\ x=\dfrac{7}{12}\)

\(g,\dfrac{x-3}{15}=\dfrac{-2}{5}\\ 5\left(x-3\right)=-30\\ x-3=-6\\ x=-6+3\\ x=-3\\ h,\dfrac{x}{-2}=\dfrac{-8}{x}\\ x^2=16\\ x=\pm\sqrt{16}\\ x=\pm4\\ k,\dfrac{x+2}{3}=\dfrac{x-4}{5}\\ 5\left(x+2\right)=3\left(x-4\right)\\ 5x+10=3x-12\\ 5x-3x=-12-10\\ 2x=-22\\ x=-11\)

\(m,\left(2x-1\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

2:

a: Xét ΔABD và ΔACD có

AB=AC

BD=CD

AD chung

=>ΔABD=ΔACD

b: ΔABD=ΔACD

=>góc ADB=góc ADC=180/2=90 độ

=>AD vuông góc BC

c: BC=12

=>BD=CD=6

AD=căn 10^2-6^2=8

d: BN=AB/2

CM=AC/2

mà AB=AC

nên BN=CM

Xét ΔNBC và ΔMCB có

NB=MC

góc NBC=góc MCB

BC chung

=>ΔNBC=ΔMCB

=>NC=BM

e: Xét ΔABC có

BM,CN là trung tuyến

BM cắt CN tại G

=>G là trọng tâm

=>A,G,D thẳng hàng và AG=2/3AD=16/3