7x-x=521.519+3.32.7
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Ta có $\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=2$
$=>2M=(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15})(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15})$
$=>2M=\sqrt{x^2-7x+19}^2-\sqrt{x^2-7x+15}^2$
$=>2M=(x^2-7x+19)-(x^2-7x+15)=4$
$=>M=2$
\(2.M=\left(x^2-7x+19\right)-\left(x^2-7x+15\right)=4\Rightarrow M=2\)
Đặt \(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=B\) = B
Xét tích \(AB=\left(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}\right)\left(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15}\right)\)
\(=x^2-7x+19-\left(x^2-7x+15\right)=x^2-7x+19-x^2+7x-15\)
\(=4\)
Mà \(B=2\Leftrightarrow A=2\)
Bài 12:
a) \(\left(\dfrac{1}{2}x+4\right)^2\)
\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)
\(=\dfrac{1}{4}x^2+4x+16\)
b) \(\left(7x-5y\right)^2\)
\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)
\(=49x^2-70xy+25y^2\)
c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)
\(=\left(y^2+6x^2\right)\left(y^2-6x^2\right)\)
\(=y^4-36x^4\)
d) \(\left(x+2y\right)^2\)
\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)
\(=x^2+4xy+4y^2\)
e) \(\left(x-3y\right)\left(x+3y\right)\)
\(=x^2-\left(3y\right)^2\)
\(=x^2-9y^2\)
f) \(\left(5-x\right)^2\)
\(=5^2-2\cdot5\cdot x+x^2\)
\(=25-10x+x^2\)
\(\left(7x+3\right)^2-\left(7x-1\right)\left(7x-3\right)=-12\)
\(\Rightarrow49x^2+42x+9-\left(49x^2-21x-7x+3\right)=-12\)
\(\Rightarrow70x+18=0\) \(\Rightarrow x=-\dfrac{18}{70}=-\dfrac{9}{35}\)
Với x = 6 ta có
A= 65 - 7.64 + 7.63 - 7.62 + 7.6 - 1
= 65 - (6+1).64 + (6+1).63 - (6+1).62 + (6+1).6 - 1
= 65 - 65 - 64 + 64 + 63 - 63 - 62 + 62 + 6 - 1
= 5
Tớ biết làm đúng 100%:
\((x\cdot1+x\cdot\frac{7}{9})\left(x\cdot1+x\cdot\frac{7}{20}\right)...\left(x\cdot1+x\cdot\frac{7}{9200}\right)=\frac{186}{25}\)
\(x\cdot\left(1+\frac{7}{9}\right)\cdot x\left(1+\frac{7}{20}\right)\cdot...\cdot x\left(1+\frac{7}{9200}\right)=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)(\frac{16}{9}+\frac{27}{20}+...+\frac{9207}{9200})=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)\left(\frac{2\cdot8}{1\cdot9}+\frac{3\cdot9}{2\cdot10}+...+\frac{93\cdot99}{92\cdot100}\right)=\frac{186}{25}\)
\(x^{92}\cdot\frac{2\cdot8\cdot3\cdot9\cdot...\cdot93\cdot99}{1\cdot9\cdot2\cdot10\cdot...\cdot92\cdot100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{\left(2\cdot3\cdot...\cdot93\right)\cdot\left(8\cdot9\cdot...\cdot99\right)}{\left(1\cdot2\cdot...\cdot92\right)\cdot\left(9\cdot10\cdot...\cdot100\right)}=\frac{186}{25}\)
\(x^{92}\cdot\frac{93\cdot8}{100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{186}{25}=\frac{186}{25}\)
\(x^{92}=\frac{186}{25}:\frac{186}{25}\)
\(x^{92}=1\Rightarrow x=1\)
cô tớ giải rồi . x=1 (đúng 100%)
7x.(2+x)-7x.(x+3)=14
7x.[(2+x)-(x+3)]=14
7x.[x+2-x-3]=14
7x.(-1) =14
7x =14:(-1)
7x =-14
x =-14:7
x =-2
Chúc bn học tốt
\(7x\cdot\left(2+x\right)-7x\cdot\left(x+3\right)=14\)
\(7x\cdot\left(2+x-x-3\right)=14\)
\(7x\cdot\left(-1\right)=14\)
\(7x=-14\)
\(x=-2\)
\(7x.\left(2+x\right)-7x.\left(x+3\right)=14\)
\(7x.\left(2+x-x-3\right)=14\)
\(7x.\left(-1\right)=14\)
\(7x=14:\left(-1\right)\)
\(7x=-14\)
\(x=\left(-14\right):7\)
\(x=-2\)
7x.( 2 + x ) - 7x( x + 3 ) = 14
<=> 7x( 2 + x - x - 3 ) = 14
<=> 7x.(-1) = 14
<=> -7x = 14
<=> x = -2