c: -62+x=-14+46

=>\(x-62=46-14=32\)

=>\(x=32+62=94\)

d: \(25+\left(x-5\right)=-123-\left(15-123\right)\)

=>\(x-5+25=-123-15+123\)

=>\(x+20=-15\)

=>\(x=-15-20=-35\)

`c) -62 + x = -14 + 46`

`-62 + x = 32`

`x = 32 - (-62)`

`x = 94`

Vậy `x = 94`

`d) 25 + ( x - 5 ) = -123 - ( 15 - 123 )`

`25 + ( x - 5 ) = -123 - 15 + 123`

`25 + ( x - 5 ) = - 15`

`x - 5 = -15 - 25`

`x - 5 = -40`

`x = -40 + 5`

`x = -35`

Vậy `x = -35`

123 + [ 54 + ( - 123 ) + 46 ]

= 123 + [ - 69 + 46 ]

= 123 + ( - 23 )

= 100 

TK

Bài làm

123 + [ 54 + ( -123 ) + 46 ]

= 123 + [ 54 - 123 + 46 ]

= 123 - 23

= 100

123 + [54 + (-123) + 46]

=123+54+(-123)+46

=[123+(-123)]+(54+46)

=0+100

=100

B) 2x + (-416) - (231 - 416 )

= 2x - 416 - 231 + 416

= 2x - ( 416 +231 - 416 )

= 2x - 231

D) 211-(123-x)+(123-211)

= 211 - 123 + x + 123 - 211

= ( 211 - 123 +123 - 211 )+ x

= 0+ x

= x

b) 2x + (-461) - (231 - 416)

= 2x - 461 - (231 - 416)

= 2x - 461 + 185

= 2x - 276

d) 211 - (123 - x) + (123 - 211)

= 211 - 123 + x + 123 - 211

= (211 - 211) + (-123 + 123) + x

= 0 + 0 + x

= x

đúng 

ko sai 

10 điểm ko về chỗ

\(a,100.123-123.100\)

\(=100.\left(123-123\right)\)

\(=100.0\)

\(=0\)

\(b,10.30+70.10\)

\(=10.\left(30+70\right)\)

\(=10.100\)

\(=1000\)

a) 100 x 123 - 123 x 100

= 100 x ( 123 - 123 )

= 100 x 0

=100

b) 10 x 30 + 70 x 10

= 10 x ( 30 + 70)

= 10 x 10

= 100

sorry, dap so cua cau a là 0, mk đánh nhầm số

TK:

\(1,\\ a,=5\sqrt{2}-12\sqrt{2}+10\sqrt{2}=3\sqrt{2}\\ b,=\dfrac{\sqrt{5}-2}{5-4}-\dfrac{\sqrt{5}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}=\sqrt{5}-2-\sqrt{5}=-2\\ c,=3\sqrt{3}-\left|1-\sqrt{3}\right|-\dfrac{6\sqrt{3}}{3}=3\sqrt{3}-\left(\sqrt{3}-1\right)-2\sqrt{3}=1\)

\(2,\\ a,P=\left(\dfrac{1}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}\right)\left(1+\dfrac{2\left(\sqrt{a}-2\right)}{a-2\sqrt{a}}\right)\left(a>0;a\ne4\right)\\ P=\dfrac{\sqrt{a}+2+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\left[1+\dfrac{2\left(\sqrt{a}-2\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}\right]\\ P=\dfrac{2\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\sqrt{a}+2}{\sqrt{a}}=\dfrac{2}{\sqrt{a}-2}\)

\(b,\sqrt{4x+12}-3\sqrt{x+3}+\sqrt{16x+48}=9\left(x\ge-3\right)\\ \Leftrightarrow2\sqrt{x+3}-3\sqrt{x+3}+4\sqrt{x+3}=9\\ \Leftrightarrow3\sqrt{x+3}=9\Leftrightarrow\sqrt{x+3}=3\\ \Leftrightarrow x+3=9\Leftrightarrow x=6\left(tm\right)\)

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