Bai 1:
√x + 1 + √4 - x + √( x + 1) ( 4 - x ) = (Dk - 1 ≤ x ≤ 4).
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\(\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(x+1\right)\left(4-x\right)}=5\)
<=> \(2\sqrt{x+1}+2\sqrt{4-x}+2\sqrt{\left(x+1\right)\left(4-x\right)}=10\) (*)
Dat: \(\sqrt{x+1}+\sqrt{4-x}=a\) \(\left(a\ge0\right)\)
=> \(a^2-5=2\sqrt{\left(x+1\right)\left(4-x\right)}\)
Khi đó pt (*) trở thành:
\(2a+a^2-5=5\)
<=> \(a^2+2a-10=0\)
Đến đây tự giải tiếp, k giải đc ib mk
Bài 2 : 1 + ( -2 ) + 3 + ( -4 ) + ... + 2015
= [ 1 + ( -2 ) ] + [ 3 + ( -4 ) ] + ... + 2015
= -1 + -1 + ... + 2015
Có số các cặp số bằng ( -1 ) là :
2014 : 2 = 1007 ( cặp )
= -1007 + 2015
= 1008
a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)
=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)
=>\(6x-\dfrac{39}{4}=1\)
=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)
=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)
b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)
=>\(2x-6=3x+6-x+1\)
=>2x-6=2x+7
=>-6=7(vô lý)
c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)
=>\(x^2+3x+x^2-2x=2x^2-2x\)
=>3x-2x=-2x
=>3x=0
=>x=0
d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)
=>\(3x^2-3x-2x-4-2x=x^2-x\)
=>\(3x^2-7x-4-x^2+x=0\)
=>\(2x^2-6x-4=0\)
=>\(x^2-3x-2=0\)
=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)
1.a
1/2+1/4+1/8+1/16+1/32
= 1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32
= 1-1/32=31/32
1b
\(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3} +\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{6}+\frac{2}{3}+\frac{1}{4}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{5}{20}+\frac{5}{30}+\frac{20}{30}+\frac{5}{20}+\frac{1}{20}+\frac{1}{30}\)
\(=\left(\frac{5}{20}+\frac{5}{20}+\frac{1}{20}\right)+\left(\frac{5}{30}+\frac{20}{30}+\frac{1}{30}\right)\)
\(=\frac{11}{20}+\frac{26}{30}\)
\(=\frac{11}{20}+\frac{13}{15}\)
\(=\frac{17}{12}\)
BÀI 1 :
a) \(-\frac{5}{8}=\frac{x}{16}\)
\(\Rightarrow x=\frac{5.16}{-8}=\frac{80}{-8}=-10\)
b) \(\frac{y}{10}=-\frac{4}{8}\)
\(\Rightarrow y=\frac{-4.10}{8}=-\frac{40}{8}=-5\)
bài 8
1) \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\)
\(\frac{x}{3}=-\frac{5}{6}+\frac{1}{4}\)
\(\frac{x}{3}=-\frac{7}{12}\)
\(\Rightarrow x=-\frac{7.3}{12}=-\frac{21}{12}=-\frac{7}{4}\)
2) \(x+\frac{3}{15}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{3}{15}\)
\(x=\frac{8}{15}\)
3) \(x-\frac{12}{4}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{12}{4}\)
\(x=\frac{7}{2}\)
4) \(\frac{3}{4}x=\frac{1}{2}\)
\(x=\frac{3}{4}:\frac{1}{2}\)
\(x=\frac{3}{2}\)
\(\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(x+1\right)\left(4-x\right)}=5\)
<=> \(2\sqrt{x+1}+2\sqrt{4-x}+2\sqrt{\left(x+1\right)\left(4-x\right)}=10\) (*)
Dat: \(\sqrt{x+1}+\sqrt{4-x}=a\) \(\left(a\ge0\right)\)
=> \(a^2-5=2\sqrt{\left(x+1\right)\left(4-x\right)}\)
Khi đó pt (*) trở thành:
\(2a+a^2-5=5\)
<=> \(a^2+2a-10=0\)
Đến đây tự giải tiếp, k giải đc ib mk