12x(x-5)+7x(3-x)=5
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-12x(x - 5) + 7x(3 - x) = 5
-12x2 + 60 + 21x - 7x2 = 5
-19x2 + 21x + 60 - 5 = 0
-19x2 + 21x + 55 = 0
\(\Delta=b^2-4ac=21^2-4.55.\left(-19\right)=441+4180=4621>0\)
vậy phtrinh có hai nghiệm phân biệt:
\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-21+\sqrt{4621}}{2.\left(-19\right)}=\frac{-21+\sqrt{4621}}{-38}\)
\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-21-\sqrt{4621}}{2.\left(-19\right)}=\frac{-21-\sqrt{4621}}{-38}\)
\(-12x\left(x-5\right)+7x\left(3-x\right)=5\)
\(-12x^2+60+21x-7x^2=5\)
\(-19x^2+60+21x=5\)
\(-19x^2+21x=5-60=-55\)
\(-19x^2+21x+55=0\)
\(a=-19;b=-21;c=55\)
\(=B^2-4ac\)
\(=21^2-4\left(-19\right).55\)
\(\Delta=4621\)
Giá trị delta cao hơn 0, vì vậy pt có hai nghiệm
\(x_1=\frac{-b-\sqrt{\Delta}}{2a};x_2=\frac{-b+\sqrt{\Delta}}{2a}\)
\(x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{\left(-21\right)-\sqrt{4621}}{2.\left(-19\right)}=\frac{-21-\sqrt{4621}}{-38}\)
\(x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-21+\sqrt{4621}}{2.\left(-19\right)}=\frac{-21+\sqrt{4621}}{-38}\)
\(-12.\left(x-5\right)+7.\left(3-x\right)=5\)
\(\Rightarrow-12x+60+21-7x=5\)
\(\Rightarrow-12x-7x=5-60-21\)
\(\Rightarrow-19x=-76\)
\(\Rightarrow x=4\)
Vậy x = 4
-12x(X-5)+7x(3-X)=5-12x(X-5)+7x(3-X)=5-12x(X-5)+7x(3-X)=5
\(\Leftrightarrow21x-7x^2-12x^2+60x=0\Leftrightarrow-19x^2+81x=0\)
\(\Leftrightarrow x\left(-19x+81\right)=0\Leftrightarrow x=0;x=\dfrac{81}{19}\)
Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
a: \(2^{x^2-2x+1}=1\)
=>\(2^{\left(x-1\right)^2}=2^0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
b: \(7^{x^2+7x}=5764801\)
=>\(7^{x^2+7x}=7^8\)
=>\(x^2+7x=8\)
=>\(x^2+7x-8=0\)
=>(x+8)(x-1)=0
=>\(\left[{}\begin{matrix}x+8=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)
c: \(6^{x^2+12x}=6^{7x}\)
=>\(x^2+12x=7x\)
=>\(x^2+5x=0\)
=>x(x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
d: \(\left(\dfrac{1}{3}\right)^{x-1}=3^{2x-5}\)
=>\(3^{-x+1}=3^{2x-5}\)
=>-x+1=2x-5
=>-x-2x=-5-1
=>-3x=-6
=>x=2
e: \(\left(\dfrac{1}{5}\right)^{3x+5}=5^{2x+1}\)
=>\(5^{-3x-5}=5^{2x+1}\)
=>-3x-5=2x+1
=>-5x=6
=>\(x=-\dfrac{6}{5}\)
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
-12.(x-5)+7.(3-x)=5
=>-12x-(-60)+21-7x=5
=>-12x+60+21-7x=5
=>-12x-7x=5-21-60
=>(-12-7).x=-76
=>-19x=-76
=>x=(-76):(-19)
=>x=4
`G(x)+H(x)=(21x^2+1+17x)+(-2+6x^3-12x^2-8)`
`=21x^2+1+17x-2+6x^3-12x^2-8`
`= 6x^3+(21x^2-12x^2)+17x+(1-2-8)`
`= 6x^3+9x^2+17x-9`
`G(x)-H(x)=(21x^2+1+17x)-(-2+6x^3-12x^2-8)`
`= 21x^2+1+17x+2-6x^3+12x^2+8`
`= -6x^3+(21x^2+12x^2)+17x+(1+2+8)`
`= -6x^3+33x^2+17x+11`
`----`
`M(x)+N(x)=(7x^5 + 1 + 17x^4 - 2)+(6x^4 - 12x^2 - 23x^4 + x)`
`= 7x^5 + 1 + 17x^4 - 2+6x^4 - 12x^2 - 23x^4 + x`
`= 7x^5+(17x^4+6x^4-23x^4)-12x^2+x+(1-2)`
`= 7x^5-12x^2+x-1`
`M(x)-N(x)=(7x^5 + 1 + 17x^4 - 2)-(6x^4 - 12x^2 - 23x^4 + x)`
`= 7x^5 + 1 + 17x^4 - 2-6x^4 + 12x^2 + 23x^4 - x`
`= 7x^5+(17x^4-6x^4+23x^4)+12x^2-x+(1-2)`
`= 7x^5+34x^4+12x^2-x-1`
12x( x - 5 ) + 7x( 3 - x ) = 5
<=> 12x2 - 60x + 21x - 7x2 - 5 = 0
<=> 5x2 - 39x - 5 = 0 (?)
Δ = b2 - 4ac = (-39)2 + 4.5.5 = 1521 + 100 = 1621
Δ > 0 nên pt có hai nghiệm phân biệt : x1 = (39+√1621)/10 ; x2 = (39-√1621)/10
Vậy ...
lậy chúa tôi lớp 6 học pt bậc 2 ạ :))