chứng minh:
(7x+1)^2- (x+7)^2=48 (x^2-1)
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Xét \(\left(7x+1\right)^2-\left(x+7\right)^2-48\left(x^2-1\right)\)
\(=49x^2+14x+1-x^2-14x-49-48x^2+48\)
\(=0\)
Vậy \(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)
Bài 1:
\(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow\left(4x-4x+5\right)\left(4x+4x-5\right)=15\)
\(\Leftrightarrow5\left(8x-5\right)=15\)
\(\Leftrightarrow8x=8\Leftrightarrow x=1\)
Vậy x = 1
Bài 2:
\(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)
\(=\left(7x+1-x-7\right)\left(7x+1+x+7\right)\)
\(=\left(6x-6\right)\left(8x+8\right)\)
\(=48\left(x-1\right)\left(x+1\right)\)
\(=48\left(x^2-1\right)=VP\)
\(\Rightarrowđpcm\)
x2-4x+7 = 0 ⇔ x2 -4x + 4 + 3 = 0
⇔ (x-2)2+3=0 ⇔ (x-2)2=-3 (vô lí)
Vậy pt vô nghiệm
*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm
Ta có: \(x^2-4x+7=0\)
\(\Leftrightarrow x^2-4x+4+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3=0\)
mà \(\left(x-2\right)^2+3\ge3>0\forall x\)
nên \(x\in\varnothing\)(đpcm)
\(\left(7x+1\right)^2-\left(x+7\right)^2=\left(7x+1\right).\left(7x+1\right)-\left(x+7\right)\left(x+7\right)=\left(49x^2+7x+7x+1\right)-\left(x^2+7x+7x+49\right)\)\(=49x^2+14x+1-x^2-7x-7x-49=\left(49x^2-x^2\right)+\left(14x-7x-7x\right)-\left(49-1\right)=48x^2-48=48.\left(x^2-1\right)\)
(7x - 1)2 - (x+7)2 = 48(x2 - 1)
<=> 49x2 + 14x + 1 - x2 - 14x - 49 = 48x2 - 48
<=>48x2 - 48 = 48x2 - 48
<=> 0x = 0(luôn đúng)
Vậy ĐPCM
1. a) 7x2 - 5x - 2 = 7x2 - 7x + 2x - 2 = 7x(x - 1) + 2(x - 1) = (x - 1).(7x + 2)
2. 5(2x - 1)2 - 3(2x - 1) = 0
<=> (2x - 1).[5(2x - 1) - 3] = 0
<=> (2x - 1).(10x - 8) = 0
<=> (2x - 1) = 0 hoặc (10x - 8) = 0
<=> x = 1/2 hoặc x = 4/5
3. x2 - 4x + 7 = (x2 - 4x + 4) + 3 = (x - 2)2 + 3
Do: (x - 2)2 > hoặc = 0 (với mọi x)
Nên (x - 2)2 + 3 > hoặc = 3 (với mọi x)
Hay (x - 2)2 + 3 > 0 (với mọi x) => đpcm
`4)x^2-5x+6`
`=x^2-2x-3x+6`
`=x(x-2)-3(x-2)=(x-2)(x-3)`
`5)x^2+7x+10`
`=x^2+5x+2x+10`
`=x(x+5)+2(x+5)=(x+5)(x+2)`
`6)x+7\sqrt{x}+10` `ĐK: x >= 0`
`=(\sqrt{x})^2+5\sqrt{x}+2\sqrt{x}+10`
`=\sqrt{x}(\sqrt{x}+5)+2(\sqrt{x}+5)=(\sqrt{x}+5)(\sqrt{x}+2)`
`7)3x^4+7x^2+4`
`=3x^4+3x^2+4x^2+4`
`=3x^2(x^2+1)+4(x^2+1)=(x^2+1)(3x^2+4)`
`8)x^2-x-2`
`=x^2-2x+x-2`
`=x(x-2)+(x-2)=(x-2)(x+1)`
`9)x^6-x^3-2`
`=x^6+x^3-2x^3-2`
`=x^3(x^3+1)-2(x^3+1)`
`=(x^3+1)(x^3-2)`.
\(M=\left(x-1\right)^3-x\left(x+2\right)^2+7x\left(x+\dfrac{1}{7}\right)\)
\(M=x^3-3x^2+3x-1-x\left(x^2+4x+4\right)+7x^2+x\)
\(M=x^3-2x^2+3x-1-x^3-4x^2-4x+7x^2+x\)
\(M=\left(x^3-x^3\right)+\left(-2x^2-x^2-4x^2+7x^2\right)+\left(3x-4x+x\right)-1\)
\(M=-1\)
Vậy...........
Chúc bạn học tốt!!!
Theo đầu bài ta có:
\(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)
\(\Rightarrow\left[\left(7x+1\right)+\left(x+7\right)\right]\left[\left(7x+1\right)-\left(x+7\right)\right]=\left(7^2-1^2\right)\left(x^2-1^2\right)\)
\(\Rightarrow\left(8x+8\right)\left(6x-6\right)=\left[\left(7+1\right)\left(7-1\right)\right]\left[\left(x+1\right)\left(x-1\right)\right]\)
\(\Rightarrow8\left(x+1\right)\cdot6\left(x-1\right)=8\left(x+1\right)\cdot6\left(x-1\right)\)( đpcm )