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Nhiều vậy sao giải @@
a) Đặt \(a=\sqrt{1+x}+\sqrt{8-x}\)
\(\Leftrightarrow a^2=1+x+8-x+2\sqrt{\left(1+x\right)\left(8-x\right)}\)
\(\Leftrightarrow a^2=9+2\sqrt{\left(1+x\right)\left(8-x\right)}\)
\(\Leftrightarrow\frac{a^2-9}{2}=\sqrt{\left(1+x\right)\left(8-x\right)}\)
\(pt\Leftrightarrow a+\frac{a^2-9}{2}=3\)
\(\Leftrightarrow\frac{a^2+2a-9}{2}=3\)
\(\Leftrightarrow a^2+2a-9=6\)
\(\Leftrightarrow a^2+2a-15=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-5\end{matrix}\right.\)
Tới đây thay vào rồi tìm x
b) \(2\left(x^2+2\right)=5\sqrt{x^3+1}\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\)
Ta có : \(a^2+b^2=x^2-x+1+x+1=x^2+2\)
\(pt\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2+2b^2-5ab=0\)
\(\Leftrightarrow2a^2-4ab+2b^2-ab=0\)
\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)
Tới đây thay vào rồi lại giải tiếp
p/s: Mình bận rồi, bao giờ rảnh giải tiếp
a) + \(VT=\sqrt{x^2+2x+10}+x^2+2x+1+7\)
\(=\sqrt{x^2+2x+1}+\left(x+1\right)^2+7>0\forall x\)
=> ptvn
d) ĐK : \(x^2+7x+7\ge0\)
Đặt \(t=\sqrt{x^2+7x+7}\ge0\) \(\Rightarrow t^2=x^2+7x+7\)
\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)
\(\Leftrightarrow3t^2+2t-5=0\Leftrightarrow\left(3t+5\right)\left(t-1\right)=0\)
\(\Leftrightarrow t=1\) ( do \(3t+5>0\forall t\ge0\) )
\(\Leftrightarrow x^2+7x+1=0\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) ( TM )
f) ĐK : \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\) thì pt trở thành :
\(a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
Lời giải:
Để biểu thức có nghĩa thì:
a) \(-7x\geq 0\Leftrightarrow x\leq 0\)
b) \(8-x\geq 0\Leftrightarrow x\leq 8\)
c) \(3x+11\geq 0\Leftrightarrow 3x\geq -11\Leftrightarrow x\geq \frac{-11}{3}\)
d) \(\frac{2x}{5}\geq 0\Leftrightarrow x\geq 0\)
e) \(-7x+5\geq 0\Leftrightarrow 5\geq 7x\Leftrightarrow x\leq \frac{5}{7}\)
f) \(\frac{1}{-2+x}\geq 0\Leftrightarrow -2+x>0\Leftrightarrow x-2>0\Leftrightarrow x>2\)
g) \(2+x^2\geq 0\) :Luôn đúng với mọi $x$ do \(x^2\geq 0\Rightarrow x^2+2\geq 2>0\)
h) \(\left\{\begin{matrix} x+7\geq 0\\ x-8\geq 0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x\geq -7\\ x\geq 8\end{matrix}\right.\Rightarrow x\geq 8\)
i) \((x+2)(x-3)\geq 0\)
\(\Leftrightarrow \left[\begin{matrix} x+2\geq 0; x-3\geq 0\\ x+2\leq 0; x-3\leq 0\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x\geq -2; x\geq 3\\ x\leq -2; x\leq 3\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x\geq 3\\ x\leq -2\end{matrix}\right.\)
k) \(\left\{\begin{matrix} \frac{x+5}{3-x}\geq 0\\ 3-x\neq 0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x+5\geq 0; 3-x>0\\ x+5\leq 0; 3-x< 0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x\geq -5; x<3 \\ x\leq -5; x>3(\text{vô lý})\end{matrix}\right.\)
\(\Rightarrow 3> x\geq -5\)
Câu a:
ĐKXĐ:...........
\(\sqrt{x^2-x+9}=2x+1\)
\(\Rightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-x+9=(2x+1)^2=4x^2+4x+1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+5x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x(x-1)+8(x-1)=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (x-1)(3x+8)=0\end{matrix}\right.\Rightarrow x=1\)
Vậy.....
Câu b:
ĐKXĐ:.........
Ta có: \(\sqrt{5x+7}-\sqrt{x+3}=\sqrt{3x+1}\)
\(\Rightarrow (\sqrt{5x+7}-\sqrt{x+3})^2=3x+1\)
\(\Leftrightarrow 5x+7+x+3-2\sqrt{(5x+7)(x+3)}=3x+1\)
\(\Leftrightarrow 3(x+3)=2\sqrt{(5x+7)(x+3)}\)
\(\Leftrightarrow \sqrt{x+3}(3\sqrt{x+3}-2\sqrt{5x+7})=0\)
Vì \(x\geq -\frac{7}{5}\Rightarrow \sqrt{x+3}>0\). Do đó:
\(3\sqrt{x+3}-2\sqrt{5x+7}=0\)
\(\Rightarrow 9(x+3)=4(5x+7)\)
\(\Rightarrow 11x=-1\Rightarrow x=\frac{-1}{11}\) (thỏa mãn)
Vậy..........
I) xd mọi x
\(\sqrt{x^2-8x+16}+\sqrt{x^2-10x+25}=9\)
\(\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-5\right)^2}=9=>\left|x-4\right|+\left|x-5\right|=9\)
\(\left[{}\begin{matrix}x< 4\Rightarrow4-x+5-x=>x=0\left(n\right)\\4\le x< 5\Rightarrow x-4+5-x=9\left(vn\right)\\x\ge5\Rightarrow x-4+x-5=9\Rightarrow x=9\left(n\right)\\\end{matrix}\right.\)
kết luận
\(\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
d)
ĐK: $x\leq \frac{16}{7}$
PT $\Rightarrow 16-7x=11^2=121$
$\Rightarrow 7x=16-121=-105$
$\Leftrightarrow x=-15$ (thỏa mãn)
e) ĐK: $x\geq 3$
PT $\Rightarrow 10(x-3)=30$ (bình phương 2 vế)
$\Leftrightarrow x-3=3\Leftrightarrow x=6$
(thỏa mãn)
f)
ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{x-2}=6$
$\Rightarrow x-2=6^2=36\Leftrightarrow x=38$ (thỏa mãn)
a)
ĐK: $x\geq \frac{-5}{2}$
PT $\Rightarrow 2x+5=25$ (bình phương 2 vế)
$\Leftrightarrow 2x=10\Leftrightarrow x=5$ (thỏa mãn)
b) ĐK: $x\geq \frac{-1}{3}$
PT $\Rightarrow 3x+1=10$ (bình phương 2 vế)
$\Leftrightarrow 3x=9\Leftrightarrow x=3$ (thỏa mãn)
c)
ĐK: $x\geq 7$
PT $\Leftrightarrow \sqrt{x-7}=3+0=3$
$\Rightarrow x-7=3^2$
$\Leftrightarrow x=16$ (thỏa mãn)