a,  \(\overline{xx}+x+5=125\)

      \(x\times11+x=120\)

      \(x\left(1+11\right)=120\)

      \(x=120\div12\)

      \(x=10\)

  Vậy x = 10

\(\overline{xxx}+\overline{xx}+x+x=992\)

\(x\times111+x\times11+x\times2=992\)

\(x\left(111+11+2\right)=992\)

 \(x=992\div124\)      

\(x=8\)

 Vậy x = 8

\(4725+\overline{xxx}+\overline{xx}+x=54909\)

\(x\times111+x\times11+x=50184\)

\(x\left(111+11+1\right)=50184\)

\(x\times123=50184\)

\(x=408\)

Vậy x = 408

b,  \(\overline{xxx}-\overline{xx}-x-25=4430\)

\(x\times111-x\times11-x=4455\)

\(x\left(111-11-1\right)=4455\)

\(x=4455\div99\)

\(x=45\)

Vậy x = 45

\(\overline{xxx}+\overline{xx}+x+x+x+1=1001\)

\(x\times111+x\times11+x\times3=1000\)

\(x\left(111+11+3\right)=1000\)

\(x=1000\div125\)

\(x=8\)

Vậy x = 8

\(35655-\overline{xxx}-\overline{xx}-x=5274\)

\(x\times111+x\times11+x=30381\)

\(x\left(111+11+1\right)=30381\)

\(x=30381\div123\)\

\(x=247\)

 Vậy \(x=247\)

a) Ta có BH//CF mà CF _|_ AB nên BH _|_ AB

Xét \(\Delta ABH\)vuông tại B có BE là đường cao nên \(AB^2=AH\cdot AE\Rightarrow AC^2=AH\cdot AE\)(vì AE=AC)

b) Vẽ DK _|_ AB khi đó DK là đường trung bình của \(\Delta FBC\)

\(\Rightarrow DK=\frac{1}{2}CF\)

tam giác ABD vuông tại A, DK là đường cao nên \(\frac{1}{DK^2}=\frac{1}{DB^2}+\frac{1}{DA^2}\)

Do đó\(\frac{1}{\left(\frac{CF}{2}\right)^2}=\frac{1}{\left(\frac{BC}{2}\right)^2}+\frac{1}{DA^2}\Rightarrow\frac{4}{CF^2}=\frac{4}{BC^2}+\frac{1}{AD^2}\)

\(\Rightarrow\frac{1}{CF^2}=\frac{1}{BC^2}+\frac{1}{4AD^2}\)

Điều kiện để biểu thức có nghĩa:

\(\hept{\begin{cases}\frac{7x-1}{2x^2+3}\ge0\\3x-2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}7x-1\ge0\\3x-2\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}7x\ge1\\3x\ge2\end{cases}}\Rightarrow\hept{\begin{cases}x\ge\frac{1}{7}\\x\ge\frac{2}{3}\end{cases}}\Rightarrow x\ge\frac{2}{3}\)

Vậy \(x\ge\frac{2}{3}\) thì BT A có nghĩa

Mình làm tắt thôi nhé

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

Dễ thấy \(A\ge0\)

\(A=\frac{x^4-2x^2+1}{x^4+x^3+x+1}=\frac{x^4-2x^3+x^2+2x^3-4x^2+2x+x^2-2x+1}{x^4-x^3+x^2+2x^2-2x^2+2x+x^2-x+1}\)

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

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

\(=\frac{x^2-2x+1}{x^2-x+1}\)

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

Ta có : \(\frac{\left(x-1\right)^2}{x^2-x+1}=\frac{\left(x-1\right)^2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge0\)

=> Đpcm

\(A=\left(\frac{1}{3}-1\right)\left(\frac{1}{6}-1\right)\left(\frac{1}{10}-1\right)....\left(\frac{1}{36}-1\right)\)

\(=\frac{-2}{3}.\frac{-5}{6}.\frac{-9}{10}...\frac{-35}{36}\)

\(=-\left(\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{70}{72}\right)=-\left(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{7.10}{8.9}\right)\)

\(=-\left(\frac{1.4.2.5.3.6...7.10}{2.3.3.4.4.5...8.9}\right)=-\frac{\left(1.2.3.4..7\right)\left(4.5.6...10\right)}{\left(2.3.4...8\right)\left(3.4.5...9\right)}=-\frac{1.10}{8.3}=\frac{-10}{24}\)

Đặt biểu thức là A .

Ta có :

\(A=\left(\frac{1}{3}-1\right)\left(\frac{1}{6}-1\right)\left(\frac{1}{10}-1\right)...\left(\frac{1}{36}-1\right)\)

\(A=\frac{-2}{3}\cdot\frac{-5}{6}\cdot\frac{-9}{10}\cdot....\cdot\frac{-36}{36}\)

\(A=-\left(\frac{4}{6}\cdot\frac{10}{12}\cdot\frac{18}{20}\cdot...\cdot\frac{70}{72}\right)\)

\(A=-\left(\frac{1.4}{2.3}\cdot\frac{2.5}{3.4}\cdot\frac{3.6}{4.5}...\frac{7.10}{8.9}\right)\)

\(A=-\left(\frac{1.4.2.5.3.6...7.10}{2.3.3.4.4.5...8.9}\right)\)

\(A=-\frac{\left(1.2.3.4....7\right)\left(4.5.6....10\right)}{\left(2.3.4....8\right)\left(3.4.5....9\right)}\)

\(A=-\frac{1.10}{8.3}\)

\(A=\frac{-10}{24}\)

a) \(3x-18.28:14=308\)

\(=>3x-36=308\)

\(=>3x=344\)\(\)

\(=>x=\frac{344}{3}\)

\(\)b)\(38+\left|x\right|=\left(-12\right)+65\)

\(=>38+\left|x\right|=53=>\left|x\right|=15=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)

c)\(15+3.\left(x-1\right):5=30\)

\(=>15+3.\left(x-1\right)=150\)

\(=>3.\left(x-1\right)=135\)

\(x-1=45=>x=46\)

d) \(5.\left(7-3x\right)+7.\left(2+2x\right)=1\)

\(=>35-15x+14+14x=1\)

\(=>49-x=1=>x=48\)

e)\(3.\left(x-7\right)=21\)

\(=>x-7=7=>x=14\)

g)\(\left[\left(6x-72\right):2-84\right]:28=5628\)

\(\left(6x-72\right):2-84=157584\)

\(\left(6x-72\right):2=157668\)

\(6x-72=315336\)

\(6x=315408=>x=52568\)

h)\(123-5\left(x+4\right)=28\)

\(5.\left(x+4\right)=95\)

\(x+4=19=>x=15\)

p)\(14.\left(x-3\right)-138=73\)

\(14.\left(x-3\right)=211\)

\(x-3=\frac{211}{14}=>x=\frac{253}{14}\)

\(t\)

\(\left|x\right|-5=2\)

\(\left|x\right|=7=>\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

v)\(4x-72=\left|100-8.25\right|\)

\(4x-72=\left|-100\right|\)

\(4x-72=100=>4x=172=>x=43\)

cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^

Ai giúp em đi ạ

Em hứa sẽ vote nhiệt tình luôn huhuuu

a, (x² - 3)(2 + 4x)

= 2x² + 4x³ - 6 - 12x

b, 6(3x + 5)(x - 4)

= (18x + 30)(x - 4)

= 18x² - 72 + 30x - 120

= 18x² + 30x - 192

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

\(\Leftrightarrow x^4+x^2+1+2x^3+2x^2+2x=y^2-y+1\)

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

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

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

\(\Leftrightarrow\left(2x^2+2x-2y+3\right)\left(2x^2+2x+2y+1\right)=3\)

Đến đây chắc khó.

Còn 10 con

'mà bươi' là mười ba 

=13 con gà đập chết 3 con thì còn 10 con.

Đúng không  bẹn

ĐKXĐ: \(x>0\)

Ta có: \(P\sqrt{x}=\left(\sqrt{x}+1\right)^2\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow x-4\sqrt{x}+4+\sqrt{x-4}=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)

Vì \(\left(\sqrt{x}-2\right)^2\ge0;\sqrt{x-4}\ge0\forall x\)

\(\Rightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}\ge0\forall x\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x}-2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow x=4\) ( tm )

Vậy...

a) \(\left(a-b\right)^2=3\)\(\Rightarrow a^2-2ab+b^2=3\)

mà \(a^2+b^2=8\)\(\Rightarrow8-2ab=3\)

\(\Rightarrow2ab=5\)\(\Rightarrow ab=\frac{5}{2}\)

Vậy \(ab=\frac{5}{2}\)

b) Ta có: \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

mà \(a-b=2\)và \(a+b=4\)

\(\Rightarrow a^2-b^2=2.4=8\)

Vậy \(a^2-b^2=8\)

a) Ta có: \(\hept{\begin{cases}a^2+b^2=8\\\left(a-b\right)^2=3\end{cases}}\Leftrightarrow\hept{\begin{cases}a^2+b^2=8\\a^2-2ab+b^2=3\end{cases}}\)

=> \(a^2+b^2-\left(a^2-2ab+b^2\right)=8-3\)

<=> \(2ab=5\)

=> \(ab=\frac{5}{2}\)

b) Ta có: \(a^2-b^2=\left(a-b\right)\left(a+b\right)=2.4=8\)

lm lộn đề nên hơi chậm xíu^^