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

Toán Lớp 7: B1: 1/ 125³ : 9³ 2/ (0,125)³ . 512 3/ (0,25)⁴. 1024 B2: a/ 3 – (-6/7)⁰ + (1/2)² : 2

Toán Lớp 7: B1:
1/ 125³ : 9³ 2/ (0,125)³ . 512 3/ (0,25)⁴. 1024
B2:
a/ 3 – (-6/7)⁰ + (1/2)² : 2 b/ (-2)³ + 2² + (-1)²⁰ + (-2)⁰
c/ ((3)²)² – ((-5)²)² + ((-2)³)² d/ 2⁴+8.[(-2)² : 1/2]⁰ – 2^-2 .4+(-2)²
e/ 2³+3.(1/2)⁰ – 2^-2.4+[(-2)² : 1/2].8
B3:
a) (x-1)^x+2=(x-1)^x+6
b) (x+20)¹⁰⁰ +|y+4|=0
e cần gấp ạ

Comments ( 1 )

  1. Giải đáp: + Lời giải và giải thích chi tiết:
    Bài $1$.
    $1$/ 125^3 : 9^3 = {1953125}/{729}
    $2$/ (0,125)^3 . 512 =  (1/8)^3 . 512 = 1/{512} . 512 = 1
    $3$/ (0,25)^4 . 1024 = (1/4)^4 . 2^{10} = 1/{2^{8}} . 2^{10} = 2^2 = 4
    Bài $2$. 
    $a$/ 3 – (-6/7)^0 + (1/2)^2 : 2
    = 3 – 1 + 1/8
    = 2 + 1/8
    = {17}/8.
    $b$/ (-2)^3 + 2^2 + (-1)^{20} + (-2)^ 0
    = -8 + 4 + 1 + 1
    = -4 + 2
    = -2.
    $c$/ $(3^2)^2 – [(-5)^2)^2] + [(-2)^3]^2$
    $= 3^4 – 5^4 + 2^6$
    $= 81 – 625 + 64$
    $= -480$.
    $d$/ 2^4 + 8.[(-2)^2 : 1/2]^0 – 2^{-2} . 4 + (-2)^2
    = 2^4 + 8.1 – 1/4 . 4 + 4
    = 16 + 8 – 1 + 4
    = 27.
    $e$/ 2^3 + 3. (1/2)^0 – 2^{-2} . 4 + [(-2)^2 : 1/2] . 8
    = 8 + 3.1 – 1/4 . 4 + (4 : 1/2) . 8
    = 8 + 3 – 1 + 64
    = 74.
    Bài $3$.
    $a$) $(x-1)^{x+2} = (x-1)^{x+6}$
    $⇔ (x-1)^{x+2} – (x-1)^{x+6} = 0$
    $⇔ (x-1)^{x+2} [1 – (x-1)^4] = 0$
    $⇒$ \(\left[ \begin{array}{l}x-1=0\\(x-1)^4=1\end{array} \right.\) 
    $⇔$ \(\left[ \begin{array}{l}x=1\\x=2\\x=0\end{array} \right.\) 
        Vậy $x$ $∈$ {0;1;2}.
    $b$) $(x+20)^{100} + |y+4| = 0$
    Vì $(x+20)^{100} + |y+4|$  $≥$ $0$  $∀$  $x;y$
    $⇒$ $(x+20)^{100} + |y+4| = 0$ khi $\begin{cases} (x+20)^{100}=0\\|y+4|=0 \end{cases}$
    $⇒$ $(x+20)^{100} + |y+4| = 0$ khi $\begin{cases} x=-20\\y=-4 \end{cases}$
       Vậu (x;y)=(-20;-4).

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

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