tim x
9(4x+3)2=16(3x-5)2
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a: \(\Leftrightarrow9x^2-12x+4-6x^2-16x=0\)
\(\Leftrightarrow3x^2-28x+4=0\)
\(\text{Δ}=\left(-28\right)^2-4\cdot3\cdot4=736>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{28-4\sqrt{46}}{6}=\dfrac{14-2\sqrt{46}}{3}\\x_2=\dfrac{14+2\sqrt{46}}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow16x^2+24x+9-16x^2+25=12\)
=>24x+34=12
=>24x=-22
hay x=-11/12
⇔(2 x ) 2+2.2 x .1+1 2=0 ... c ,(3 x −4) 2−14(3 x −4)(6+3 x )+49(3 x +6)=16 ... ⇔9 x 2−24 x +16−126 x 2−252 x +168 x +336+147 x +294=16.
https://olm.vn/hoi-dap/detail/192758180810.html
a) x- 18 = - 25
x =( -25 ) + 18
x = -7
b) 42 - x = 53 - 60
16 - x = 65
x = 16 - 65
x = -49
a: B=A(5x+3)
=(3x^3-4x+1)(5x+3)
=15x^4+9x^3-20x^2-12x+5x+3
=15x^4+9x^3-20x^2-7x+3
b: \(B=\dfrac{3x^4+6x^3-4x^2-7x+2}{3x^3-4x+1}\)
\(=\dfrac{3x^4-4x^2+x+6x^3-8x+2}{3x^3-4x+1}\)
=x+2
\(\left(\dfrac{3x}{4}+5\right)-\left(\dfrac{2x}{3}-4\right)-\left(\dfrac{x}{6}+1\right)=\left(\dfrac{1}{3}+4\right)-\left(\dfrac{1}{3}x-3\right)\)
\(\Leftrightarrow\dfrac{3x}{4}-\dfrac{2x}{3}-\dfrac{x}{6}+5+4-1=\dfrac{13}{3}-\dfrac{1}{3}x+9\)
\(\Leftrightarrow\dfrac{9x-8x-2x}{12}+8=\dfrac{13-x}{3}+\dfrac{27}{3}\)
\(\Leftrightarrow\dfrac{-x}{12}+\dfrac{96}{12}=\dfrac{40-x}{3}\Leftrightarrow\dfrac{96-x}{12}=\dfrac{160-4x}{12}\)
\(\Rightarrow96-160=-4x+x\Leftrightarrow-64=-3x\Leftrightarrow x=\dfrac{64}{3}\)
\(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)
\(\Rightarrow9\left(16x^2+24x+9\right)=144x^2-480x+400\)
\(\Rightarrow696x=319\)
\(\Rightarrow x=319:696\)
\(\Rightarrow x=\frac{11}{24}\)
\(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)
=> \(3^2\left(4x+3\right)^2-4^2\left(3x-5\right)^2=0\)
=> \(\left(12x+9\right)^2-\left(12x-20\right)^2=0\)
=> \(\left(12x+9-12x-20\right)\left(12x+9+12x+20\right)=0\)
=> \(-11\left(24x+29\right)=0\)
=> x = -29/24
Chúc bạn làm bài tốt