1.9: Các số là căn bậc hai số học của 49 là: \(\sqrt{\left(-7\right)^2}\)
Câu 1.8:
a:
ĐKXĐ: x>=0
\(\sqrt{x}=3\)
=>\(x=3^2\)
=>x=9(nhận)
b: ĐKXĐ: x>=0
\(\sqrt{x}=\sqrt5\)
=>x=5(nhận)
c: ĐKXĐ: x>=0
\(\sqrt{x}=0\)
=>x=0(nhận)
d: ĐKXĐ: x>=0
\(\sqrt{x}=-2\)
mà \(\sqrt{x}\ge0\forall x\) thỏa mãn ĐKXĐ
nên x∈∅
Bài 1.7:
a: \(x^2=25\)
=>\(\left[\begin{array}{l}x=5\\ x=-5\end{array}\right.\)
b: \(x^2=30,25\)
=>\(x^2=5,5^2\)
=>x=5,5 hoặc x=-5,5
c: \(x^2=5\)
=>\(\left[\begin{array}{l}x=\sqrt5\\ x=-\sqrt5\end{array}\right.\)
d: \(x^2-\sqrt3=\sqrt2\)
=>\(x^2=\sqrt3+\sqrt2\)
=>\(x=\pm\sqrt{\sqrt3+\sqrt2}\)
e: \(x^2-5=0\)
=>\(x^2=5\)
=>\(\left[\begin{array}{l}x=\sqrt5\\ x=-\sqrt5\end{array}\right.\)
f: \(x^2+\sqrt5=2\)
mà \(x^2+\sqrt5\ge\sqrt5>2\forall x\)
nên x∈∅
g: \(x^2=\sqrt3\)
=>\(\left[\begin{array}{l}x=\sqrt[4]{3}\\ x=-\sqrt[4]{3}\end{array}\right.\)
h: \(2x^2+3\sqrt2=2\sqrt3\)
=>\(2x^2=2\sqrt3-3\sqrt2=\sqrt{12}-\sqrt{18}<0\) (vô lý)
=>x∈∅
i: \(\left(x-1\right)^2=1\frac{9}{16}\)
=>\(\left(x-1\right)^2=\frac{25}{16}\)
=>\(\left[\begin{array}{l}x-1=\frac54\\ x-1=-\frac54\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac54+1=\frac94\\ x=-\frac54+1=-\frac14\end{array}\right.\)
j: \(x^2=\left(1-\sqrt3\right)^2\)
=>\(\left[\begin{array}{l}x=1-\sqrt3\\ x=-\left(1-\sqrt3\right)=\sqrt3-1\end{array}\right.\)
k: \(x^2=27-10\sqrt2\)
=>\(x^2=\left(5-\sqrt2\right)^2\)
=>\(\left[\begin{array}{l}x=5-\sqrt2\\ x=-5+\sqrt2\end{array}\right.\)
l: \(x^2+2x=3-2\sqrt3\)
=>\(x^2+2x+1=4-2\sqrt3\)
=>\(\left(x+1\right)^2=\left(\sqrt3-1\right)^2\)
=>\(\left[\begin{array}{l}x+1=\sqrt3-1\\ x+1=-\sqrt3+1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt3-2\\ x=-\sqrt3\end{array}\right.\)
