`a)1/7xx2/7+1/7xx5/7+6/7`

`=1/7xx(2/7+5/7)+6/7`

`=1/7xx1+6/7`

`=1/7+6/7=1`

`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`

`=6/11xx(4/9+7/9-2/9)`

`=6/11xx9/9`

`=6/11`

Sorry nãy ghi thiếu.

`c)4/25xx5/8xx25/4xx24`

`=(4xx5xx25xx24)/(25xx8xx4)`

`=(4xx5xx24)/(4xx8)`

`=(5xx24)/8`

`=5xx3=15`

a, \(\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}+\dfrac{6}{7}\)

\(=\dfrac{1}{7}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.1+\dfrac{6}{7}\)

\(=\dfrac{1}{7}+\dfrac{6}{7}=1\)

b, \(\dfrac{6}{11}.\dfrac{4}{9}+\dfrac{6}{11}.\dfrac{7}{9}-\dfrac{6}{11}.\dfrac{2}{9}\)

\(=\dfrac{6}{11}.\left(\dfrac{4}{9}+\dfrac{7}{9}-\dfrac{2}{9}\right)\)

\(=\dfrac{6}{11}.1=\dfrac{6}{11}\)

c, \(\dfrac{4}{25}.\dfrac{5}{8}.\dfrac{25}{4}.24\)

\(=\left(\dfrac{4}{25}.\dfrac{25}{4}\right).\left(\dfrac{5}{8}.24\right)\)

\(=1.15=15\)

\(=\dfrac{2}{3}+\dfrac{1}{5}-\dfrac{2}{3}-4\)

\(=\dfrac{1}{5}-4=\dfrac{-19}{5}\)

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)

Do đó: x=18; y=12; z=9

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)

Do đó: x=18; y=12; z=9

a) Thay x + 3y - 2z vào biểu thức ta có:

 \(\dfrac{x - 1}{3} = \dfrac{3(y + 2)}{3 . 4} = \dfrac{2(z - 2)}{2 . 3}\) = \(​​​​\dfrac{x - 1}{3} = \dfrac{3x + 6}{12} = \dfrac{2z - 4}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhua ta có:

\(​​​​\dfrac{x - 1}{3} = \dfrac{3y + 6}{12} = \dfrac{2z - 4}{6} = ​​​​\dfrac{x - 1}{3}+ \dfrac{3y + 6}{12} -\dfrac{2z - 4}{6}\) 

=\(​​​​\dfrac{x - 1 + 3y + 6 - 2z + 4}{3 + 12 -6} \) = \(​​​​\dfrac{(x + 3y - 2z) + ( -1 + 6 +4)}{3 + 12 - 6} \)

=\(​​​​\dfrac{36 + 9}{9}\) = 5

=> \(​​​​\dfrac{x - 1}{3} =\) 5 => x - 1 = 5.3 =15 => x = 5+1 = 6

=>

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Vậy ...

(Bạn dựa theo cách này và lm những bài tiếp nhé!)

cái jz nè .-. 

cái này mà là toán lớp 6 á ;-;

a: VT=sin^2a(sin^2a+cos^2a)+cos^2a

=sin^2a+cos^2a

=1=VP

b: \(VT=\dfrac{sina+sina\cdot cosa+sina-sina\cdot cosa}{1-cos^2a}=\dfrac{2sina}{sin^2a}=\dfrac{2}{sina}=VP\)

c: \(VT=\dfrac{sin^2a+1+2cosa+cos^2a}{sina\left(1+cosa\right)}\)

\(=\dfrac{2\left(cosa+1\right)}{sina\left(1+cosa\right)}=\dfrac{2}{sina}=VP\)

đặt

\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+..+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot97}\)

\(3A=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot97}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{97}\\ 3A=\dfrac{95}{194}\\ A=\dfrac{95}{582}\)

\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)

\(\left(4-x\right)^2=25=5^2=\left(-5\right)^2\)

4-x=5 hoặc 4-x=-5

x=-1 hoặc x=9