2

Ta có:

3a+2b-c-d=1 (1)

2a+2b-c+2d=2 (2)

4a-2b-2c+d=3 (3)

8a+b-6c+d=4 (4)

(1)+(2)+(3)-(4) vế theo vế ta được:

a+b+c+d=1+2+3-4=2

Vâp a+b+c+d=2

CÁC BẠN GIÚP MÌNH NHA!

đề bài là j vậy

\(b=a+b+c+d-\left(a+c+d\right)=1-2=-1\\ c=a+b+c+d-\left(a+b+d\right)=1-3=-2\\ d=a+b+c+d-\left(a+b+c\right)=1-4=-3\\ a=a+b+c+d-b-c-d=1+1+2+3=7\)

b = a + b + c + d − ( a + c + d ) = 1 − 2 = − 1

c = a + b + c + d − ( a + b + d ) = 1 − 3 = − 2

d = a + b + c + d − ( a + b + c ) = 1 − 4 = − 3

a = a + b + c + d − b − c − d = 1 + 1 + 2 + 3 = 7

Circle the word that has the underlined part pronounced differently from the others

1A closure B leisure C pleasure D sure

2A contury B cotnbine C collect D concert

3A future B return C picture D culture

4A general B garden C go D get

1a 2d 3b 4d 5b

1A. eighteen B. handbag C. waiting D. practice

2A. transport (n) B. enter (v) C. cancel (v) D. report (v)

3A. passenger B. horrible C. suggestion D. goalkeeper

4A. hesitate B. merchandise C. quantify D. disappear

5A. drinkable B.possessive C. pilotless D. driverless

circle the word with a different strss pattern from others

1a national b physical c arrival d natural

2a classical b ponsonous c logical d pollution

3a nature b classic c degree d debris

4a dramatic b tornado c historic d infury

5a examinee b electronic c scientific d preparation

6a industry b bilology c natural d musical

7a geography b economic c scientific d preparation

8a debris b rainstorm c destroy d shelter

9a climatology b bibliography c communication d radiography

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a-c}{3b-d}=\dfrac{3bk-dk}{3b-d}=k\)

\(\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=k\)

Do đó: \(\dfrac{3a-c}{3b-d}=\dfrac{2a+3c}{2b+3d}\)

c: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)

\(\dfrac{2ab+b^2}{2cd+d^2}=\dfrac{2\cdot bk\cdot b+b^2}{2\cdot dk\cdot d+d^2}=\dfrac{b^2}{d^2}\)

Do đó: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{2ab+b^2}{2cd+d^2}\)

Tớ lỡ tay ấn nhầm, làm tiếp nhá.

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a+2c}{3b+2d}\) (ĐPCM).

c) Ta có:

+) \(\dfrac{a}{c}=\dfrac{b}{d}\) mà \(\dfrac{b}{d}=\dfrac{2b}{2d}\)

\(\Rightarrow\dfrac{a}{c}=\dfrac{2b}{2d}\)

Áp dụng TCDTSBN, ta có:

\(\Rightarrow\dfrac{a}{c}=\dfrac{2b}{2d}=\dfrac{a-2b}{c-2d}\) (ĐPCM)

d) Ta có:

+) \(\dfrac{a}{c}=\dfrac{b}{d}\) mà \(\dfrac{a}{c}=\dfrac{5a}{5b};\dfrac{b}{d}=\dfrac{2b}{2d}\)

\(\Rightarrow\dfrac{5a}{5c}=\dfrac{2b}{2d}\)

Áp dụng TCDTSBN, ta có:

\(\Rightarrow\dfrac{5a}{5c}=\dfrac{2b}{2d}=\dfrac{5a-2b}{5c-2d}\) (ĐPCM)

ĐPCM là điều phải chứng minh nhá bạn, còn áp dụng TCDTSBN là áp dụng tính chất dãy tỉ số bằng nhao

Chúc bạn học tốt!

a) Ta có:

+) \(\dfrac{a}{b}=\dfrac{c}{d}\) mà \(\dfrac{c}{d}=\dfrac{4c}{4d}\)

\(\Rightarrow\)\(\dfrac{a}{b}=\dfrac{4c}{4d}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{b}=\dfrac{4c}{4d}=\dfrac{a+4c}{b+4d}\)(ĐPCM)

b) Ta có:

+) \(\dfrac{a}{b}=\dfrac{c}{d}\) mà \(\dfrac{a}{b}=\dfrac{3a}{3b}\); \(\dfrac{c}{d}=\dfrac{2c}{2d}\)

\(\Rightarrow\) \(\dfrac{3a}{3b}=\dfrac{2c}{2d}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có: