3(2x-1)-5(x-3)+6(3x-4) = 24

<=>6x-3-5x+15+18x-24=24

<=>19x-12=24

<=>19x=36

<=>x=36/19

vậy....

3(2x-1)-5(x-3)+6(3x-4) = 24

<=>6x-3-5x+15+18x-24=24

<=>19x-12=24

<=>19x=36

<=>x=36/19

bài này bạn đã đưa lên và   đã có người  giải rồi mà

\(x^2-13x+40=0\)

\(\Leftrightarrow x^2-5x-8x+40=0\)

\(\Leftrightarrow x\left(x-5\right)-8\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=8\end{cases}}}\)

Vậy ......

x²-13x+40=0

<=>x²-5x-8x+40=0

<=>x(x-5)-8(x-5)=0

<=>(x-5)(x-8)=0

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=8\end{cases}}\)

vậy...

\(\left(2x-1\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\3x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=1\\3x=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-2}{3}\end{cases}}\)

Để (2x—1).(3x+2)=0

Thì 2x—1=0 hoặc 3x+2=0

==>2x=0+1 hoặc 3x=0–2

==>2x=1 hoặc 3x=—2

==>  x=1:2 hoặc x=—2:3

==> x=1/2 hoặc x=—2/3

Có \(\left(a+b\right)^2\ge0\Rightarrow a^2+2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge-2ab\Leftrightarrow-\left(a^2+b^2\right)\ge2ab\)

\(a^3+b^3+2\ge2ab+a+b\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)+2-2ab-a-b\ge0\)

\(\Leftrightarrow2-2ab-\left(a+b\right)\ge0\)

\(\Leftrightarrow2-2ab\ge0\)

\(\Leftrightarrow2--\left(a^2+b^2\right)\ge0\Leftrightarrow2+a^2+b^2\ge0\)

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4/ a/ Ta có \(x^2-2xy+y^2+a^2=\left(x-y\right)^2+a^2\)

Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\a^2\ge0\end{cases}}\)=> \(\left(x-y\right)^2+a^2\ge0\)

=> \(x^2-2xy+y^2+a^2\ge0\)

Vậy \(x^2-2xy+y^2\)chỉ nhận những giá trị không âm.

b/ Ta có \(x^2+2xy+2y^2+2y+1=\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x+y\right)^2+\left(y+1\right)^2\)

Mà \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)=> \(\left(x+y\right)^2+\left(y+1\right)^2\ge0\)

=> \(x^2+2xy+2y^2+2y+1\ge0\)

Vậy \(x^2+2xy+2y^2+2y+1\)chỉ nhận những giá trị không âm.

c/ Ta có \(9b^2-6b+4c^2+1=\left(3b-1\right)^2+4c^2\)

Mà \(\hept{\begin{cases}\left(3b-1\right)^2\ge0\\4c^2\ge0\end{cases}}\)=> \(\left(3b-1\right)^2+4c^2\ge0\)

=> \(9b^2-6b+4c^2+1\ge0\)

Vậy \(9b^2-6b+4c^2+1\)chỉ nhận những giá trị không âm.

d/ Ta có \(x^2+y^2+2x+6y+10=\left(x+1\right)^2+\left(y+3\right)^2\)

Mà \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)=> \(\left(x+1\right)^2+\left(y+3\right)^2\ge0\)

=> \(x^2+y^2+2x+6y+10\ge0\)

Vậy \(x^2+y^2+2x+6y+10\)chỉ nhận những giá trị không âm.

1/

a/ \(x^4-y^4=\left(x^2-y^2\right)\)

b/ \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

                                                  \(=2b\left[a^2+2ab+b^2-\left(a^2-b^2\right)+\left(a^2-2ab+b^2\right)\right]\)

                                                  \(=2b\left(a^2+b^2\right)\)

c/ \(\left(a^2+2ab+b^2\right)+\left(a+b\right)\)

= \(\left(a+b\right)^2+\left(a+b\right)\)

= \(\left(a+b\right)\left(a+b+1\right)\)

A = \(\frac{1}{2}.24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^{32}-1\right)\left(5^{32}+1\right)\)

A = \(\frac{1}{2}\left(5^{64}-1\right)\)

\(2A=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

        \(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

       \(=5^{64}-1\)

=> \(A=\frac{5^{64}-1}{2}\)

Ta có; \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\Rightarrow\widehat{A}+\widehat{B}=220^o\)

Lại có: \(\left(\widehat{A}+\widehat{B}\right)+\left(\widehat{A}-\widehat{B}\right)=220^o+10^o\)

\(\Rightarrow2\widehat{A}=230^o\Rightarrow\widehat{A}=115^o\)

\(\Rightarrow B=115^o-10^o=105^o\)

Vậy..