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a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)
b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)
c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)
\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)
d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)
\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)
a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)
<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)
<=> \(\sqrt{x}+8=28\)
<=> \(\sqrt{x}=28-8\)
<=> \(\sqrt{x}=20\)
<=> \(\left(\sqrt{x}\right)^2=20^2\)
<=> x = 400
=> x = 400
b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)
<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)
<=> \(3\sqrt{x}+5=\sqrt{x}+12\)
<=> \(3\sqrt{x}=\sqrt{x}+12-5\)
<=> \(3\sqrt{x}=\sqrt{x}+7\)
<=> \(3\sqrt{x}-\sqrt{x}=7\)
<=> \(2\sqrt{x}=7\)
<=> \(\sqrt{x}=\frac{7}{2}\)
<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)
<=> \(x=\frac{49}{4}\)
=> \(x=\frac{49}{4}\)
c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)
<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)
<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)
<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)
<=> \(8\sqrt{x}=6\sqrt{x}+4\)
<=> \(8\sqrt{x}-6\sqrt{x}=4\)
<=> \(2\sqrt{x}=4\)
<=> \(\sqrt{x}=2\)
<=> \(\left(\sqrt{x}\right)^2=2^2\)
<=> x = 4
=> x = 4
d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)
<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)
<=>\(2\sqrt{3x}=6\sqrt{3x}\)
<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)
<=>\(-4\sqrt{3x}=0\)
<=> \(\sqrt{3x}=0\)
<=> \(\left(\sqrt{3x}\right)^2=0^2\)
<=> 3x = 0
<=> x = 0
=> x = 0
a, 3 : ( 1 - 3/2x ) = 4 : ( 2 - x )
<=> \(\frac{3}{1-\frac{3}{2}x}=\frac{4}{2-x}\)
<=> 3 ( 2 - x ) = 4 ( 1 - 3/2x )
<=> 6 - 3x = 4 - 6x
<=> -3x + 6x = 4 - 6
<=> 3x = -2
<=> x = -2/3
b, 2.3x + 3x-1 = 7( 32 + 2.62 )
b, 2.3x + 3x-1 = 7( 32 + 2.62 )
<=> 2.3x + 3x-1 = 7.81
<=> 3x-1(2.3 + 1) = 7.81
<=> 3x-1.7 = 7.81
<=> 3x-1=81
<=> 3x-1 = 34
=> x - 1 = 4 => x = 5
a) \(\sqrt{x-1}=5\)
\(\Leftrightarrow x-1=25\)
\(\Rightarrow x=26\)
b)\(\sqrt{\left(x-\frac{1}{3}\right)^2}=7\)
\(\Leftrightarrow x-\frac{1}{3}=7\)
\(\Rightarrow x=\frac{22}{3}\)
c)\(\sqrt{x+1}+5=3\)
làm tương tự nha bạn
P/s tham khảo nha
a) \(\sqrt{x-1}=5\Leftrightarrow\left(\sqrt{x-1}\right)^2=5^2\)
\(\Leftrightarrow\sqrt{x-1}=25\)
\(\Leftrightarrow x=25+1=26\)
b) \(\sqrt{\left(x-\frac{1}{3}^2\right)}=7\). Đơn giản hóa phép tính:
\(\sqrt{\left(x-\frac{1}{3}\right)^2}\)với \(x-\frac{1}{3}\)
\(\Rightarrow x-\frac{1}{3}=7\)
\(x=7+\frac{1}{3}\Leftrightarrow x=\frac{22}{3}\)
c) \(\sqrt{1+x}+5=3\)
\(\sqrt{1-x}=3-5\)
\(\sqrt{1-x}=-2\)
\(\Leftrightarrow1+x=4\)
\(x=4-1=3\)
Mở rộng thêm:
When \(x=3\) the original equation \(\sqrt{1+x}+5=3\) does not hold true.
We will drop \(x=3\) from the solution set. (tự dịch nha! Vì mình sử dụng chương trình để trợ giúp mình giải
Ta cố bdt \(|a|+|b|\ge|a+b|\), dễ dàng chứng mình bằng bình phương 2 vế. Dấu = sảy ra <=>IaI.IbI=a.b <=> a.b>=0
áp dụng vào từng câu
a)A=Ix+1I+Ix+2I+Ix+3I+I-x-4I+I-x-5I ( vì Ix+4I=I-x=4I, Ix+5I=I-x-5I
A>=I(x+1)+(-x-5)I+I(x+2)+(-x-4)I +Ix+3I=4+2+Ix+3I=6+Ix+3I>=6
Dấu bằng khi (x+1)(-x-5)>=0;(x+2)(-x-4)>=0;Ix+3I=0 =>x=-3
b) LÀm tương tự MinB=18
Dấu = khi (2x+1)(-2x-11)>=0;(2x+3)(-2x-9)>=0;(2x+5)(-2x-7)>=0 <=>-7/2<=x<=-5/2
b/ \(\left|\left|3x-1+9\right|\right|=-\left(-31\right)\)
<=> \(\left|\left|3x+8\right|\right|=31\)
<=> \(\left|3x+8\right|=31\)
<=> \(\orbr{\begin{cases}3x+8=-31\\3x+8=31\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=-39\\3x=23\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-13\\x=\frac{23}{3}\end{cases}}\)
a)
<=> 7 + 9 = x^2
<=> 16 = x^2
<=> 4 = x
b)
<=> x^2 = 18 - 9
<=> x^2 = 9
<=> x = 3
a) √72 + 32 = 7 + 9 = 16 = 42 = x2
Vậy x=4
b) x2 + 9 =18
<=> x2 = 18 -9
<=> x2 = 9 = 32
Vậy x=3