\(a,B=4\sqrt{x=1}-3\sqrt{x+1}+2\)\(\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
\(b,\)đưa về \(\sqrt{x+1}=4\Rightarrow x=15\)
a, Với \(x\ge-1\)
\(\Rightarrow B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
b, Ta có B = 16 hay
\(4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\)bình phương 2 vế ta được
\(\Leftrightarrow x+1=16\Leftrightarrow x=15\)
a) B = 4√x+1 b) x = 15
) .
b) Đưa về . Suy ra .
a) \(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=\sqrt{16}.\sqrt{x+1}+\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}+\sqrt{x+1}=\sqrt{x+1}\left(4-3+2+1\right)=4\sqrt{x+1}\)
b) \(B=16\Leftrightarrow4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\Rightarrow x\in\left\{15\right\}\)
a) 4\(\sqrt{x+1}\) b)\(x=15\)
a) B= 4\(\sqrt{x-1}\)
b) x=15
a. B = \(4\sqrt{x+1}\)
b. Ta có
B = 16
⇔ \(\sqrt{x+1}\) = 4
⇒ x = 15
a) B=\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)=\(4\sqrt{x+1}\)
b)x=15
a) B = \(4\sqrt{x+1}\)
b) x=15
a/ B= 4\(\sqrt{x+1}\) - 3\(\sqrt{x+1}\) + 2\(\sqrt{x+1}\) + \(\sqrt{x+1}\) = 4\(\sqrt{x+1}\)
b/ x = 15
a. \(4\sqrt{x+1}\)
b. x=15
a)B=\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\) (với x\(\ge-1\))
B=\(\sqrt{4^2\cdot\left(x+1\right)}-\sqrt{3^2\cdot\left(x+1\right)}+\sqrt{2^2\cdot\left(x+1\right)}+\sqrt{x+1}\)
B=\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
B=\(4\sqrt{x+1}\)
Vậy B=\(4\sqrt{x+1}\) với x≥-1
b) Có B=4\(\sqrt{x+1}\) với x≥-1
Mà B = 16
⇔4\(\sqrt{x+1}\) =16
⇔\(\sqrt{x+1}=4\)
⇔x+1=16
⇔x=15(Thỏa mãn điều kiện )
Vậy x=15 thì B= 16
B=\(\sqrt{16x+16}\)- \(\sqrt{9x+9}\)+\(\sqrt{4x+4}\)+\(\sqrt{x+1}\) x\(\ge\)-1
a) B=4\(\sqrt{x+1}\)-3\(\sqrt{x+1}\)+2\(\sqrt{x+1}\)+\(\sqrt{x+1}\)
B= 4\(\sqrt{x+1}\)(TM)
Vậy với x\(\ge\)-1 thì B =4\(\sqrt{x+1}\)
b)Với x\(\ge-1\)thì B=16
4\(\sqrt{x+1}\)=16
\(\Leftrightarrow\)\(\sqrt{x+1}\)=4
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\)(TMDK)
Vậy với B=16 thì x=15
\(4\sqrt{x+1}\) b, x=15
a. B = \(\sqrt{16x+16}\) - \(\sqrt{9x+9}\) + \(\sqrt{4x+4}\) + \(\sqrt{x+1}\) với x \(\ge\)-1
B = 4\(\sqrt{x+1}\) - 3\(\sqrt{x+1}\) + 2\(\sqrt{x+1}\) + \(\sqrt{x+1}\)
B = 4\(\sqrt{x+1}\)
Vậy vs x\(\ge-1\) thì B = 4\(\sqrt{x+1}\)
b. vs x\(\ge-1\)
\(\Rightarrow\) B = 16
\(\Leftrightarrow4\sqrt{x+1}\) = 16
\(\Leftrightarrow\sqrt{x+1}\) = 4
\(\Leftrightarrow\) x+1 = 16
\(\Leftrightarrow\) x = 15 (tm)
vậy B = 16 thì x = 15
a) B=4\(\sqrt{x+1}\)-3\(\sqrt{x+1}\)+2\(\sqrt{x+1}\)+\(\sqrt{x+1}\)=4\(\sqrt{x+1}\)
b) B= 16
\(\Leftrightarrow\)4\(\sqrt{x+1}\)=16
\(\Leftrightarrow\)\(\sqrt{x+1}\)=4
\(\Leftrightarrow\)(\(\sqrt{x+1}\))\(^2\)=4\(^2\)
\(\Leftrightarrow\)x+1=16
\(\Leftrightarrow\)x=15
a) .
b) Đưa về . Suy ra .
bài 3
a) B=\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}với\ge-1\) b) B=16 theo đầu bài
B=\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\) => 16=\(4\sqrt{x+1}\)
B=\(\sqrt{x+1}\left(4-3+2+1\right)\) <=> 16/4=\(\sqrt{x+1}\)
B=\(4\sqrt{x+1}\) <=> 4=\(\sqrt{x+1}\)
<=> \(4^2=\left(\sqrt{x+1}\right)^2\)
<=> 16 = x+1
<=> 15=x
Vậy x=15
b, Để B = 16 thì
⇔ x + 1 = 16 ⇔ x = 15 (thỏa mãn x ≥ -1)
a)B=\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)
=\(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
=\(4\sqrt{x+1}\)
b) Để B=16 thì \(4\sqrt{x+1}=16\) <=>\(\sqrt{x+1}=4\)
<=>\(x+1=16\)
<=>\(x=15\)(thỏa mãn đkxđ)
\(a)B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=\sqrt{4\left(x+1\right)}-\sqrt{3\left(x+1\right)}+\sqrt{2\left(x+1\right)}+\sqrt{x+1}=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=4\sqrt{x+1}\)
\(4\sqrt{x+1}=16\leftrightarrow\sqrt{x+1}=4\leftrightarrow x+1=16\leftrightarrow x=15\) (thỏa mãn x ≥ -1)
a,4\(\sqrt{x+1}\)
b,\(x=15\)