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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Tìm x, biết a. (3x + 4) ² – (3x-1)(3x+1) = 49 b. x ² – 4x + 4 = 9.(x-2) c. x ² – 25 = 3x – 15 d. (x-1) ³ + 3(x+1) ² = (x ² – 2x + 4)(x+

Toán Lớp 8: Tìm x, biết
a. (3x + 4) ² – (3x-1)(3x+1) = 49
b. x ² – 4x + 4 = 9.(x-2)
c. x ² – 25 = 3x – 15
d. (x-1) ³ + 3(x+1) ² = (x ² – 2x + 4)(x+2)

Comments ( 2 )

  1. Giải đáp:
    $a) (3x+4)²-(3x-1).(3x+1)=49$
    $⇔ 9x²+24x+16 -9x²+1=49$
    $⇔ (9x²-9x²)+24x+(16+1)=49$
    $⇔ 24x+17=49$
    $⇔ 24x=32$
    $⇔ x= \frac{4}{3}$
    $Vậy$ $x= \frac{4}{3}$
    $b) x²-4x+4=9.(x-2)$
    $⇔ (x-2)²-9.(x-2)=0$
    $⇔ (x-2).(x-11)=0$
    $⇔ (x-2).(x-11)=0$
    $⇔$ \(\left[ \begin{array}{l}x-2=0\\x-11=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=2\\x=11\end{array} \right.\) 
    $Vậy$ $x=2$ $hoặc$ $x=11$
    $c) x²-25=3x-15$
    $⇔ (x-5).(x+5)-3.(x-5)=0$
    $⇔ (x-5).(x+5-3)=0$
    $⇔ (x-5).(x+2)=0$
    $⇔$ \(\left[ \begin{array}{l}x-5=0\\x+2=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)
    $Vậy$ $x=-5$ $hoặc$ $x=8$
    $d) (x-1)³+3.(x+1)²=(x²-2x+4).(x+2)$
    $⇔ x³-3x²+3x-1+3.(x²+2x+1)- (x-2).(x²+2x+4)=0$
    $⇔ x³-3x²+3x-1+3x²+6x+3-x³-8=0$
    $⇔ (x³-x³)+(3x²-3x²)+(3x+6x)+(3-1-8)=0$
    $⇔ 9x-6=0$
    $⇔ 9x=6$
    $⇔ x= \frac{2}{3}$
    $Vậy$ $x= \frac{2}{3}$
    $@FESTIVALS$

  2. Giải đáp: a. x=$\frac{4}{3}$ 
     b. x∈{2; 11}  
    c. x=5 hoặc x=-2
    d. x=$\frac{2}{3}$ 
    Lời giải và giải thích chi tiết:
    a) $(3x+4)^{2}$-(3x-1)(3x+1)=49
         ($9x^{2}$+24x+16)-(9$x^{2}$-1)=49
          9$x^{2}$+24x+16-9$x^{2}$+1=49
           24x+17=49
            24x=32
             x=$\frac{4}{3}$ 
    b) $x^{2}$-4x+4=9(x-2)
         ($x^{2}$-4x+4)-9(x-2)=0
          $(x-2)^{2}$-9(x-2)=0
           (x-2)[(x-2)-9]=0
            (x-2)(x-11)=0
            ⇒$\left \{ {{x-2=0} \atop {x-11=0}} \right.$ 
            ⇒$\left \{ {{x=2} \atop {x=11}} \right.$ 
    c. $x^{2}$-25=3x-15
    ⇔($x^{2}$-25)-(3x-15)=0
    ⇔(x-5)(x+5)-3(x-5)=0
    ⇔(x-5)(x+5-3)=0
    ⇔(x-5)(x+2)=0
    ⇔\(\left[ \begin{array}{l}x-5=0\\x+2=0\end{array} \right.\)
    ⇒\(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\) 
    d. $(x-1)^{3}$+3($(x+1)^{2}$=($x^{2}$-2x+4)(x+2)
    ⇔$x^{3}$-3$x^{2}$+3x-1+3($x^{2}$+2x+1)=$x^{3}$+8
    ⇔-3$x^{2}$+3x+3$x^{2}$+6x=8+1-3
    ⇔9x=6
    ⇔x=$\frac{2}{3}$ 
     

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222-9+11+12:2*14+14 = ? ( )