Câu hỏi:

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Tự luận

1. Thực hiện phép tính (tính hợp lí nếu có thể).

a) A=3:(32)2+19.36 +0,75A = 3:{\left( {\frac{{ - 3}}{2}} \right)^2} + \frac{1}{9}.\sqrt {36}  + 0,75;          b) B=(823+12)(57332)(53+52+4)B = \left( {8 - \frac{2}{3} + \frac{1}{2}} \right) - \left( {5 - \frac{7}{3} - \frac{3}{2}} \right) - \left( {\frac{5}{3} + \frac{5}{2} + 4} \right).

2. Tìm xx, biết:

a) 34(x+12)=14\frac{3}{4} - \left( {x + \frac{1}{2}} \right) = \frac{1}{4};                             b) (x23)2+1625=1{\left( {x - \frac{2}{3}} \right)^2} + \frac{{16}}{{25}} = 1.

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Lời giải

Hướng dẫn giải:

1. a) A=3:(32)2+19  .36 +0,75A = 3:{\left( {\frac{{ - 3}}{2}} \right)^2} + \frac{1}{9}\,\,.\,\sqrt {36}  + 0,75=3:94+19.6+0,75 = 3:\frac{9}{4} + \frac{1}{9}.6 + 0,75

=3  .  49+23+0,75 = 3\,\,.\,\,\frac{4}{9} + \frac{2}{3} + 0,75A=43+23+0,75A = \frac{4}{3} + \frac{2}{3} + 0,75=2+0,75 = 2 + 0,75=2,75 = 2,75.

b) B=(823+12)(57332)(53+52+4)B = \left( {8 - \frac{2}{3} + \frac{1}{2}} \right) - \left( {5 - \frac{7}{3} - \frac{3}{2}} \right) - \left( {\frac{5}{3} + \frac{5}{2} + 4} \right).

=823+125+73+3253524 = 8 - \frac{2}{3} + \frac{1}{2} - 5 + \frac{7}{3} + \frac{3}{2} - \frac{5}{3} - \frac{5}{2} - 4

=(854)+(23+7353)+(12+3252) = \left( {8 - 5 - 4} \right) + \left( { - \frac{2}{3} + \frac{7}{3} - \frac{5}{3}} \right) + \left( {\frac{1}{2} + \frac{3}{2} - \frac{5}{2}} \right)

= 1+0+12 =  - 1 + 0 + \frac{{ - 1}}{2}=32 = \frac{{ - 3}}{2}.

2.

a) 34(x+12)=14\frac{3}{4} - \left( {x + \frac{1}{2}} \right) = \frac{1}{4}

x+12=3414x + \frac{1}{2} = \frac{3}{4} - \frac{1}{4}      

x+12=12x + \frac{1}{2} = \frac{1}{2}

x=1212x = \frac{1}{2} - \frac{1}{2}

x=0x = 0

Vậy x=0x = 0.

b) (x23)2+1625=1{\left( {x - \frac{2}{3}} \right)^2} + \frac{{16}}{{25}} = 1

(x23)2=11625{\left( {x - \frac{2}{3}} \right)^2} = 1 - \frac{{16}}{{25}}

(x23)2=925{\left( {x - \frac{2}{3}} \right)^2} = \frac{9}{{25}}

(x23)2=(35)2=(35)2{\left( {x - \frac{2}{3}} \right)^2} = {\left( {\frac{3}{5}} \right)^2} = {\left( {\frac{{ - 3}}{5}} \right)^2}

TH1: x23=35x - \frac{2}{3} = \frac{3}{5}

x=35+23x = \frac{3}{5} + \frac{2}{3}

x=1915x = \frac{{19}}{{15}}

TH2: x23=35x - \frac{2}{3} = \frac{{ - 3}}{5}

x=35+23x = \frac{{ - 3}}{5} + \frac{2}{3}

x=115x = \frac{1}{{15}}

Vậy x{1915;  115}x \in \left\{ {\frac{{19}}{{15}};\,\,\frac{1}{{15}}} \right\}.

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