Bài 1:

a: \(\left(5-\frac23+\frac37\right):\left(24\frac{4}{21}-25\frac{8}{21}\right)\)

\(=\left(\frac{105}{21}-\frac{14}{21}+\frac{9}{21}\right):\left(24+\frac{4}{21}-25-\frac{8}{21}\right)\)

\(=\frac{100}{21}:\left(-1-\frac{4}{21}\right)=\frac{100}{21}:\frac{-25}{21}=\frac{100}{-25}=-4\)

b: \(\left(2\frac46-\frac{7}{15}\right):\left(\frac23-\frac25\right)^2\)

\(=\left(2+\frac23-\frac{7}{15}\right):\left(\frac{10}{15}-\frac{6}{15}\right)^2\)

\(=\left(2+\frac{10}{15}-\frac{7}{15}\right):\left(\frac{4}{15}\right)^2=\left(2+\frac{3}{15}\right):\frac{16}{225}=\frac{11}{5}\cdot\frac{225}{16}\)

\(=\frac{11}{16}\cdot45=\frac{495}{16}\)

c: \(\left(\frac{13}{18}-\frac{1}{72}\right):\frac{1}{18}-\left(\frac49-\frac{25}{16}\right)\cdot\frac92\)

\(=\left(\frac{52}{72}-\frac{1}{72}\right)\cdot18-\left(\frac{64}{144}-\frac{100}{144}\right)\cdot\frac92\)

\(=\frac{51}{72}\cdot18+\frac{36}{144}\cdot\frac92=\frac{51}{4}+\frac14\cdot\frac92=\frac{111}{8}\)

d: \(\frac35:\left(-\frac{1}{15}-\frac16\right)+\frac35:\left(-\frac13-1\frac{1}{15}\right)\)

\(=\frac35:\left(-\frac{2}{30}-\frac{5}{30}\right)+\frac35:\left(-\frac{5}{15}-\frac{16}{15}\right)\)

\(=\frac35:\frac{-7}{30}+\frac35:\frac{-21}{15}=\frac35\cdot\frac{-30}{7}+\frac35\cdot\frac{-5}{7}=\frac35\left(-\frac{30}{7}-\frac57\right)\)

\(=\frac35\cdot\left(-5\right)=-3\)

e: \(4\left(-\frac12\right)^3-2\left(-\frac12\right)^2+3\cdot\left(-\frac12\right)+\left(-1\right)^{2002}\)

\(=4\cdot\frac{-1}{8}-2\cdot\frac14-\frac32+1\)

\(=-\frac12-\frac12-\frac32+1=-\frac32\)

f: \(\frac{2^4\cdot2^6}{\left(2^5\right)^2}-\frac{2^5\cdot15^3}{6^3\cdot10^2}\)

\(=\frac{2^{10}}{2^{10}}-\frac{2^5\cdot5^3\cdot3^3}{2^3\cdot5^3\cdot2^2\cdot5^2}=1-\frac{3^3}{5^2}=1-\frac{27}{25}=-\frac{2}{25}\)

Bài 3:

a: |x-5|=8

=>\(\left[\begin{array}{l}x-5=8\\ x-5=-8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=13\\ x=-5\end{array}\right.\)

b: \(\left|1-\frac23x\right|+\frac34-5\frac12=0\)

=>\(\left|\frac23x-1\right|+\frac34-\frac{11}{2}=0\)

=>\(\left|\frac23x-1\right|+\frac34-\frac{22}{4}=0\)

=>\(\left|\frac23x-1\right|=\frac{19}{4}\)

=>\(\left[\begin{array}{l}\frac23x-1=\frac{19}{4}\\ \frac23x-1=-\frac{19}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac23x=\frac{19}{4}+1=\frac{23}{4}\\ \frac23x=-\frac{19}{4}+1=-\frac{15}{4}\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{23}{4}:\frac23=\frac{23}{4}\cdot\frac32=\frac{69}{8}\\ x=-\frac{15}{4}:\frac23=-\frac{15}{4}\cdot\frac32=\frac{-45}{8}\end{array}\right.\)

c: |9-7x|+7=26

=>|7x-9|=26-7=19

=>\(\left[\begin{array}{l}7x-9=19\\ 7x-9=-19\end{array}\right.\Rightarrow\left[\begin{array}{l}7x=19+9=28\\ 7x=-19+9=-10\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-\frac{10}{7}\end{array}\right.\)

Câu 27: B

Câu 28: B

Câu 29: A

Câu 30: D

Câu 31: A
Câu 32: B

3: 

a: \(\Leftrightarrow x+1-6\sqrt{x+1}-9=0\)

=>\(\left(\sqrt{x+1}-3\right)=0\)

=>x+1=9

=>x=8

b: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\left(\sqrt{\dfrac{1}{2}x+1}+3\right)}}=10\)

=>\(\sqrt{\dfrac{1}{2}x-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}-\dfrac{21}{4}}=10\)

=>\(\dfrac{1}{2}x-\dfrac{21}{4}-\dfrac{7}{4}\sqrt{\dfrac{1}{2}x+1}=100\)

=>\(\dfrac{7}{4}\cdot\sqrt{\dfrac{1}{2}x+1}=\dfrac{1}{2}x-\dfrac{21}{4}-100=\dfrac{1}{2}x-\dfrac{421}{4}\)

=>\(\sqrt{\dfrac{1}{2}x+1}=\dfrac{2}{7}x-\dfrac{421}{7}\)

=>1/2x+1=(2/7x-421/7)^2

=>1/2x+1=4/49x^2-1684/49x+177241/49

=>\(x\simeq249,77;x\simeq177,36\)

Chọn D.

4.2:

a: x^2-x+1=x^2-x+1/4+3/4

=(x-1/2)^2+3/4>=3/4>0 với mọi x

=>x^2-x+1 ko có nghiệm

b: 3x-x^2-4

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

=-(x^2-3x+9/4+7/4)

=-(x-3/2)^2-7/4<=-7/4<0 với mọi x

=>3x-x^2-4 ko có nghiệm

5:

a: x^2+y^2=25

x^2-y^2=7

=>x^2=(25+7)/2=16 và y^2=16-7=9

x^4+y^4=(x^2)^2+(y^2)^2

=16^2+9^2

=256+81

=337

b: x^2+y^2=(x+y)^2-2xy

=1^2-2*(-6)

=1+12=13

x^3+y^3=(x+y)^3-3xy(x+y)

=1^3-3*1*(-6)

=1+18=19

mik cảm ơn bạn nhiều vì đã giúp mik

`7,`

`a, B+A=4x-2x^2+3`

`-> B=(4x-2x^2+3)-A`

`-> B=(4x-2x^2+3)-(x^2-2x+1)`

`B=4x-2x^2+3-x^2+2x-1`

`B=(-2x^2-x^2)+(4x+2x)+(3-1)`

`B=-3x^2+6x+2`

`b, C-A=-x+7`

`-> C=(-x+7)+A`

`-> C=(-x+7)+(x^2-2x+1)`

`-> C=-x+7+x^2-2x+1`

`C=x^2+(-x-2x)+(7+1)`

`C=x^2-3x+8`

`c,`

`A-D=x^2-2`

`-> D= A- (x^2-2)`

`-> D=(x^2-2x+1)-(x^2-2)`

`D=x^2-2x+1-x^2+2`

`D=(x^2-x^2)-2x+(1+2)`

`D=-2x+3`

`6,`

`a,`

`P+Q=4x-2x^2+3`

`-> Q=(4x-2x^2+3)-P`

`-> Q=(4x-2x^2+3)-(3x^2+x-1)`

`Q=4x-2x^2+3-3x^2-x+1`

`Q=(-2x^2-3x^2)+(4x-x)+(3+1)`

`Q=x^2+3x+4`

`b,`

`x^2-5x+2-P=H`

`-> H= (x^2-5x+2)-(3x^2+x-1)`

`H=x^2-5x+2-3x^2-x+1`

`H=(x^2-3x^2)+(-5x-x)+(2+1)`

`H=-4x^2-6x+3`

`c,`

`P-R=5x^2-3x-4`

`-> R= P- (5x^2-3x-4)`

`-> R=(3x^2+x-1)-(5x^2-3x-4)`

`R=3x^2+x-1-5x^2+3x+4`

`R=(3x^2-5x^2)+(x+3x)+(-1+4)`

`R=-2x^2+4x+3`

where do you think people will live ?

Where do you think people will live?

\(A=\dfrac{4x+2\sqrt{x}+2}{2\sqrt{x}+1}=\dfrac{2\sqrt{x}\left(2\sqrt{x}+1\right)+2}{2\sqrt{x}+1}=2\sqrt{x}+\dfrac{2}{2\sqrt{x}+1}\)

\(=2\sqrt{x}+1+\dfrac{2}{2\sqrt{x}+1}-1\ge2\sqrt{\left(2\sqrt{x}+1\right)\cdot\dfrac{2}{2\sqrt{x}+1}}-1=2\sqrt{2}-1\)

=> A \(\ge2\sqrt{2}-1\)

Dấu "=" xảy ra <=> \(2\sqrt{x}+1=\dfrac{2}{2\sqrt{x}+1}\)

<=> \(\left(2\sqrt{x}+1\right)^2=2\) <=> \(\left[{}\begin{matrix}2\sqrt{x}+1=2\\2\sqrt{x}+1=-2\left(loại\right)\end{matrix}\right.\)

<=> \(\sqrt{x}=\dfrac{1}{2}\) <=> \(x=\dfrac{1}{4}\)(tm)

Vậy minA = \(2\sqrt{2}-1\) khi x = 1/4