1.1 Hình vuông có tối đa 4 góc vậy 4 hình vuông có tối đa 20 góc. S

2.1 hình vuông có tối đa 4 góc vậy 4 hình vuông có tối đa 16 góc. Đ

3. 1 hình vuông có tối thiểu 4 góc vậy 4 hình vuông có tối thiểu 16 góc. Đ

4.1 hình vuông có tối thiểu 1 góc vậy 4 hình vuông có tối thiểu 16 góc. S

Nhiêu đó hết tài năng rồi, mình mới lớp 3 thôi.

a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)

\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)

\(=\dfrac{5}{7}\times1\)

\(=\dfrac{5}{7}\)

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

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

\(=1+0\)

\(=1\)

c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)

\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)

\(=\dfrac{5}{7}+\dfrac{2}{7}\)

\(=1\)

d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)

\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)

\(=\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+1\)

\(=\dfrac{1}{2}+1\)

\(=\dfrac{3}{2}\)

e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)

\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)

\(=\dfrac{4}{5}+\dfrac{12}{10}\)

\(=\dfrac{4}{5}+\dfrac{6}{5}\)

\(=\dfrac{10}{5}\)

\(=2\)

g) \(362\times728+326\times272\)

\(=326\times\left(728+272\right)\)

\(=326\times1000\)

\(=326000\) 

a: =91/105+60/105-101/105

=50/105=10/21

c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)

d: =2-2/9

=18/9-2/9

=16/9

e: =24/36-9/36+8/36

=23/36

g: =5/2+1/2

=3

.

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1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)

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

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

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

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)

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

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

215

215

Cái này

\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)

\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow3x+6+2x+2=5x+4\)

\(\Leftrightarrow3x+2x-5x=-6-2+4\)

\(\Leftrightarrow0x=-4\)

=> PT vô nghiệm 

\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)

\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)

\(\Leftrightarrow4x-2-15=9x-3\)

\(\Leftrightarrow4x-9x=2+15-3\)

\(\Leftrightarrow-5x=14\)

.....

mấy cái này mẫu nào dài cậu phân tích ra : 

VD : câu  3 : \(3x^2-4x+1\)

\(=3x^2-3x-x+1\)

\(=3x\left(x-1\right)-\left(x-1\right)\)

\(=\left(3x-1\right)\left(x-1\right)\)

r bắt đầu giải PHương trình :)) Mấy câu còn lại tương tự 

4; \(\frac{5}{x-2}+\frac{2}{x+4}=\frac{3x}{x^2+2x-8}.\)

\(\Leftrightarrow\frac{5\left(x+4\right)}{\left(x-2\right)\left(x+4\right)}+\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+4\right)}=\frac{3x}{\left(x-2\right)\left(x+4\right)}\)

\(\Leftrightarrow5x+20+2x-4=3x\)

\(\Leftrightarrow4x=-16\Leftrightarrow x=-2\left(TM\right)\)

KL ::

\(5;\frac{4}{x+6}+\frac{1}{x-3}=\frac{9}{x^2+3x-18}\)

\(\Leftrightarrow\frac{4\left(x-3\right)}{\left(x+6\right)\left(x-3\right)}+\frac{x+6}{\left(x-3\right)\left(x+6\right)}=\frac{9}{\left(x-3\right)\left(x+6\right)}\)

\(\Leftrightarrow4x+x=3+9-6\)

\(\Leftrightarrow5x=6\Leftrightarrow x=\frac{6}{5}\)

Bạn ơi, bạn viết lại đề đi. Khó nhìn quá

a) 1/4(x-3)+2=1/5

1/4.(x-3) = 1/5-2

1/4.(x-3) = -9/5

x-3 = (-9/5):1/4

x-3 = -36/5

x = -36/5+3

x= -21/5

\(\dfrac{1}{3}+\dfrac{5}{9}=\dfrac{6}{18}+\dfrac{10}{18}=\dfrac{16}{18}=\dfrac{8}{9}\)

\(\dfrac{1}{7}-\dfrac{1}{9}=\dfrac{9}{63}-\dfrac{7}{63}=\dfrac{2}{63}\)

\(3:\dfrac{5}{9}=3.\dfrac{9}{5}=\dfrac{27}{5}\)

\(3.\dfrac{5}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)

\(\dfrac{1}{9}.\dfrac{9}{3}=\dfrac{1}{3}\)

\(\dfrac{1}{3}:\dfrac{1}{7}=\dfrac{7}{3}\)

\(9+\dfrac{9}{3}=9+3=12\)

\(4-\dfrac{2}{4}=4-\dfrac{1}{2}=\dfrac{7}{2}\)

\(\dfrac{1}{3}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{9}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3+5}{9}\) \(=\) \(\dfrac{8}{9}\)

\(\dfrac{1}{7}\) \(-\) \(\dfrac{1}{9}\) \(=\) \(\dfrac{9}{63}\) \(-\) \(\dfrac{7}{63}\) \(=\) \(\dfrac{9-7}{63}\) \(=\) \(\dfrac{2}{63}\)

\(\dfrac{1}{9}\) \(\times\) \(\dfrac{9}{3}\) \(=\) \(\dfrac{1\times9}{9\times3}\) \(=\) \(\dfrac{1}{3}\)

\(\dfrac{1}{3}\) \(\div\) \(\dfrac{1}{7}\) \(=\) \(\dfrac{1}{3}\) \(\times\) \(\dfrac{7}{1}\) \(=\) \(\dfrac{1\times7}{3\times1}\) \(=\) \(\dfrac{7}{3}\)

\(3\) \(\div\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{1}\) \(\div\)  \(\dfrac{5}{9}\) \(=\dfrac{3}{1}\times\dfrac{9}{5}=\dfrac{3\times9}{1\times5}=\dfrac{27}{5}\)

\(3\times\dfrac{5}{9}=\dfrac{3}{1}\times\dfrac{5}{9}=\dfrac{3\times5}{1\times9}=\dfrac{5}{3}\)

\(9+\dfrac{9}{3}=\dfrac{9}{1}+\dfrac{9}{3}=\dfrac{27}{3}+\dfrac{9}{3}=\dfrac{27+9}{3}=\dfrac{36}{3}=12\)

\(4\) \(-\dfrac{2}{4}=\dfrac{4}{1}-\dfrac{2}{4}=\dfrac{16}{4}-\dfrac{2}{4}=\dfrac{14}{4}=\dfrac{7}{2}\)