a) M + (5x² - 2xy) = 6x² + 9xy - y²

=> M = (6x² + 9xy - y²) - (5x² - 2xy)

=> M = 6x² + 9xy - y² - 5x² + 2xy = x² + 11xy - y²

b) (25x²y - 13xy² + y³) - m = 11x²y - 2y³

=> m = (25x²y - 13xy²  + y³) - (11x²y - 2y³)

=> m = 25x²y - 13xy² + y³ - 11x²y + 2y³ = 14x²y - 13xy² + 3y³

c) M = 0 - (12x⁴ - 15x²y + 2xy² + 7) = -12x⁴ + 15x²y - 2xy² - 7

a,\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\) 

\(< =>M=6x^2+9xy-y^2-5x^2+2xy\)

\(< =>M=x^2+11xy-y^2\)

b,\(\left(25x^2y-13xy^2+y^3\right)-M=11x^2y-2y^3\)

\(< =>M=25x^2y-13xy^2+y^3-11x^2y+2y^3\)

\(< =>M=14x^2y-12xy^2+3y^3\)

c,\(M+\left(12x^4-15x^2y+2xy^2+7\right)=0\)

\(< =>M=15x^2y-7-2xy^2-12x^4\)

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

<=> 3( 3x - 4 ) = 2.1

<=> 9x - 12 = 2

<=> 9x = 14

<=> x = 14/9

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

3x-4.3=2.1

3x-12=2

3x=12+2

3x=14

x=14:3

x=14/3

vậy x=14/3

Bài làm:

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

=> \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=0\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=0\) (1)

Mà \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\), cách CM như sau:

\(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)

Tương tự: \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\) ; \(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\)

Cộng vế 3 BĐT trên lại ta sẽ được: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

Thay vào (1) ta được:

\(0=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\ge3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\le0\)

Dấu "=" xảy ra khi: \(a=b=c\)

nhanh nhé mik tích cho 8 cái lun

Vì AB // DM :

⇒DMAˆ=BAMˆ⇒DMA^=BAM^(2 góc so le trong)

⇒CAMˆ=EMAˆ⇒CAM^=EMA^(2 góc so le trong)

⇒DMAˆ+EMAˆ=CAMˆ+BAMˆ⇔DMEˆ=CABˆ⇒DMA^+EMA^=CAM^+BAM^⇔DME^=CAB^(1)

Vì EM // AC

⇒MECˆ=ACEˆ⇒MEC^=ACE^(2 góc so le trong)

⇒DECˆ=ECMˆ⇒DEC^=ECM^(2 góc so le trong)

⇒MECˆ+DECˆ=ACEˆ+ECMˆ⇔MEDˆ=ACMˆ⇒MEC^+DEC^=ACE^+ECM^⇔MED^=ACM^(2)

nhanh để mik tích

Đặt ab + 4 = m22 (m ∈ N)

 ⇒ab = m22− 4 = (m − 2) (m + 2)

 ⇒b =(m−2).(m+2)a(m−2).(m+2)a

Ta có:m=a+2⇒⇒ m-2=a

⇒⇒b=a(a+4)aa(a+4)a=a+4

Vậy với mọi số tự nhiên a luôn tồn tại b = a + 4 để ab + 4 là số chính phương. 

\(\frac{8^2\cdot4^5}{2^{20}}=\frac{\left(2^3\right)^2\cdot\left(2^2\right)^5}{2^{20}}=\frac{2^6\cdot2^{10}}{2^{20}}=\frac{2^{16}}{2^{20}}=2^{-4}=\frac{1}{16}\)

\(\frac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\frac{\left(3^4\right)^{11}\cdot3^{17}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^5}=\frac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{10}}=\frac{3^{61}}{3^{40}}=3^{21}\)

\(\frac{8^2.4^5}{2^{20}}=\frac{\left(2^3\right)^2.\left(2^2\right)^5}{2^{20}}=\frac{2^6.2^{10}}{2^{20}}=\frac{2^{16}}{2^{20}}=\frac{1}{2^4}=\frac{1}{16}\)

\(\frac{81^{11}.3^{17}}{27^{10}.9^{15}}=\frac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{44}.3^{17}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)

\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)

\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)

\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)