A = \(a^2\) + 2\(a^2b\) + 2\(ab^2\) + b\(^2\)

A = (\(a^2+2ab+b^2\)) - 2ab + (2\(a^2b+2ab^2\))

A = (a + b)\(^2\) + 2ab.(a+ b - 1) (1)

Thay a + b = 1 vào biểu thức (1) ta có:

A = 1\(^2\) + 2ab.(1 - 1)

A = 1 + 2.0

A = 1 + 0

A = 1

C =(a - b - c)\(^2\) - a\(^2\) - b\(^2\) - c\(^2\)

C = (a\(^{}\) - b)\(^2\) - 2(a -b)c + c\(^2\) - a\(^2\) - b\(^2\) - \(c^2\)

C = a\(^2\) - 2ab + b\(^2\) - 2ac + 2bc + c\(^2\) - \(a^2\) - \(b^2-c^2\)

C = (a\(^2\) - a\(^2\))+(\(b^2\) - b\(^2\))+(c\(^2\) - \(c^2\))-2ab - 2ac + 2bc

C = 0 + 0 + 0 - 2ab - 2ac + 2bc

C = -2ab - 2ac + 2bc

a: ĐKXĐ: x∉{4;-5}

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

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

=>\(2x^2+10x+3x+15=2x^2-8x-x+4\)

=>13x+15=-9x+4

=>22x=4-15=-11

=>\(x=-\frac{11}{22}=-\frac12\) (nhận)

b: ĐKXĐ: x∉{5;-1}

\(2-\frac{x+3}{x-5}+\frac{1-x}{x+1}=0\)

=>\(\frac{2\left(x-5\right)\left(x+1\right)}{\left(x-5\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x+1\right)}{\left(x-5\left)\left(x+1\right)\right.\right.}-\frac{\left(x-1\right)\left(x-5\right)}{\left(x-5\right)\left(x+1\right)}=0\)

=>2(x-5)(x+1)-(x+3)(x+1)-(x-1)(x-5)=0

=>\(2\left(x^2+x-5x-5\right)-\left(x^2+4x+3\right)-\left(x^2-6x+5\right)=0\)

=>\(2x^2-8x-10-x^2-4x-3-x^2+6x-5=0\)

=>-6x-18=0

=>-6x=18

=>x=-3(nhận)

c: ĐKXĐ: x∉{2;-2}

\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)

=>\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)

=>\(\frac{\left(x-2\right)^2-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)

=>\(\left(x-2\right)^2-3\left(x+2\right)=2x-22\)

=>\(x^2-4x+4-3x-6-2x+22=0\)

=>\(x^2-9x+20=0\)

=>(x-4)(x-5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\left(nhận\right)\\ x=5\left(nhận\right)\end{array}\right.\)

d: ĐKXĐ: x∉{2;-2}

Ta có: \(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)

=>\(\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)

=>\(\frac{12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

=>12-(x+1)(x+2)+(x+7)(x-2)=0

=>\(12-\left(x^2+3x+2\right)+\left(x^2-2x+7x-14\right)=0\)

=>\(12-x^2-3x-2+x^2+5x-14=0\)

=>2x-4=0

=>2x=4

=>x=2(loại)

e: ĐKXĐ: x∉{2;4}

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

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

=>(x-1)(x-4)+2=(x+3)(x-2)

=>\(x^2-5x+4+2=x^2-2x+3x-6\)

=>-5x+6=x-6

=>-6x=-12

=>x=2(loại)

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

\(n^2=4\)

\(\Rightarrow\left[\begin{array}{l}n=2\\ n=-2\end{array}\right.\)

vậy n=2 hoặc n=-2

đáp là 2 nhé bạn vì 2 mũ 2 = 4

Ta có: 100 000 000 = 99 999 999 + 1

Mà 99 999 999 ⋮ 3 và 1 chia dư 1

⇒ (99 999 999 + 1) chia 3 dư 1

⇒ 100 000 000 chia 3 dư 1

Tổng các chữ số của số 100000000 là:

1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 1

Vì 1 : 3 dư 1 nên :

100000000 : 3 dư 1

\(a.\left(5x-4\right)\left(4x+^{}6\right)=0\)

\(\left[\begin{array}{l}5x-4=0\Rightarrow x=\frac45\\ 4x+6=0\Rightarrow x=-\frac32\end{array}\right.\)

vậy x = \(\frac45\) hoặc \(x=-\frac32\)

\(b.3x^2+6x=x+2\)

\(3x\cdot\left(x+2\right)=x+2\)

\(3x\cdot\left(x+2\right)-\left(x+2\right)=0\)

\(\left(3x-1\right)\left(x+2\right)=0\)

\(\left[\begin{array}{l}3x-1=0\Rightarrow x=\frac13\\ x+2=0\Rightarrow x=-2\end{array}\right.\)

vậy x \(=\frac13\) hoặc x=-2

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

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

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

\(\left[\begin{array}{l}2x+1=0\Rightarrow x=-\frac12\\ x^2+2=0\Rightarrow x\notin O\end{array}\right.\)

vậy \(x=-\frac12\)

\(d.x^3-5x^2-4x+20=0\)

\(x^2\cdot\left(x-5\right)-4\cdot\left(x-5\right)=0\)

\(\left(x^2-4\right)\left(x-5\right)=0\)

\(\left(x-2\right)\left(x+2\right)\left(x-5\right)=0\)

\(\left[\begin{array}{l}x-2=0\Rightarrow x=2\\ x+2=0\Rightarrow x=-2\\ x-5=0\Rightarrow x=5\end{array}\right.\)

vậy x = 2 hoặc x = -2 hoặc x = 5

\(e.\left(2x+5\right)^2=16=4^2=\left(-4\right)^2\)

\(\left[\begin{array}{l}2x+5=4\Rightarrow x=-\frac12\\ 2x+5=-4\Rightarrow x=-\frac92\end{array}\right.\)

vậy \(x=-\frac12\) hoặc \(x=-\frac92\)