Ta có: \(A=\overline{5a7,34}+\overline{bc,1}+\overline{14,2d}\)

=500+10a+7+0,34-10b-c-0,1+14,2+0,01d

=10a-10b-c+0,01d+521,44

\(B=527,9+\overline{ab,cd}\overline{}\)

=527,9+10a+b+0,1c+0,01d

B-A

=527,9+10a+b+0,1c+0,01d-10a+10b+c-0,01d-521,44

=11b+1,1c+6,46>0

=>B>A

\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)

cảm ơn bạn minh nhiều nha

Bài 22:

a: \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)

b: \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)

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

\(=\left(-x-4\right)\left(2x^2+3x+4\right)\)

c: \(-16+\left(x-3\right)^2\)

\(=\left(x-3\right)^2-16\)

=(x-3-4)(x-3+4)

=(x-7)(x+1)

d: \(64+16y+y^2=y^2+2\cdot y\cdot8+8^2=\left(y+8\right)^2\)

Bài 21:

a: \(\left(\frac12+x\right)^2=x^2+2\cdot x\cdot\frac12+\left(\frac12\right)^2=x^2+x+\frac14\)

\(\left(2x+1\right)^2=\left(2x\right)^2+2\cdot2x\cdot1+1^2=4x^2+4x+1\)

b: \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)

\(\left(xy+0,01\right)^2=\left(xy\right)^2+2\cdot xy\cdot0,01+\left(0,01\right)^2\)

\(=x^2y^2+0,02xy+0,0001\)

c: \(\left(\frac12-x\right)^2=\left(\frac12\right)^2-2\cdot\frac12\cdot x+x^2=x^2-x+\frac14\)

\(\left(2x-1\right)^2=\left(2x\right)^2-2\cdot2x\cdot1+1^2=4x^2-4x+1\)

d: \(\left(2x-3y\right)^2=\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2=4x^2-12xy+9y^2\)

\(\left(xy-0,01\right)^2=\left(xy\right)^2-2\cdot xy\cdot0,01+\left(0,01\right)^2\)

\(=x^2y^2-0,02xy+0,0001\)

e: (x+1)(x-1)=x^2-1

g: (x+y+z)(x-y-z)

\(=x^2-\left(y+z\right)^2\)

\(=x^2-y^2-z^2-2yz\)

f: (x-2y)(x-2y)

\(=\left(x-2y\right)^2=x^2-4xy+4y^2\)

_Con đội Toán lớp 6 đây ạ=)))

Toán giống lớp 4

Ta có: \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}\)

            \(2y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{6}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{6}=\dfrac{x+z}{20+6}=\dfrac{52}{26}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=20.2=40\\y=15.2=30\\z=6.2=12\end{matrix}\right.\)

b)\(3x\left(x+3y\right)-6xy\left(x+3y\right)\)

\(=\left(3x-6xy\right)\left(x+3y\right)\)

c)\(x\left(x+y\right)-5x-5y\)

\(=x\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x-5\right)\left(x+y\right)\)

Bài 1: 

b. \(3x\left(x+3y\right)-6xy\left(x+3y\right)\)

= (3x - 6xy)(x + 3y)

= 3x(1 - 2y)(x + 3y)

c. \(x\left(x+y\right)-5x-5y\)

= x(x + y) - 5(x + y)

= (x - 5)(x + y)

d. \(3\left(x-y\right)-5x\left(y-x\right)\)

= 3(x - y) + 5x(x - y)

= (3 + 5x)(x - y)

Bài 3:

a. x + 6x² = 0

<=> x(1 + 6x) = 0

<=> \(\left[{}\begin{matrix}x=0\\1+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{6}\end{matrix}\right.\)

b. 2(x + 3) - x(x + 3) = 0

<=> (2 - x)(x + 3) = 0

<=> \(\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c. 5x(x - 2) - (2 - x) = 0

<=> 5x(x - 2) + (x - 2) = 0

<=> (5x + 1)(x - 2) = 0

<=> \(\left[{}\begin{matrix}5x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=2\end{matrix}\right.\)

d. (x + 1) = (x + 1)²

<=> (x + 1) - (x + 1)² = 0

<=> (1 - x - 1)(x + 1) = 0

<=> -x(x + 1) = 0

<=> \(\left[{}\begin{matrix}-x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Ta có hình vẽ:

A B C K

a/ Xét tam giác AKB và tam giác AKC có:

AK: chung

AB = AC (GT)

BK = KC (GT)

=> tam giác AKB = tam giác AKC (c.c.c)

b/ Ta có: tam giác AKB = tam giác AKC (câu a)

=> \(\widehat{AKB}\)=\(\widehat{AKC}\) (2 góc tương ứng)

Mà \(\widehat{AKB}\)+\(\widehat{AKC}\)=180⁰ (kề bù)

=> \(\widehat{AKB}\)=\(\widehat{AKC}\)=\(\frac{1}{2}\)180⁰=90⁰

=> AK vuông góc BC (đpcm)

em cần gấp lắm