Решить неравенства и у выражения

9)

4\; ,\; \; k+3-\sqrt{(4-k)^2}=k+3-|\underbrace {4-k}_{<0}|=k+3-(k-4)=7" class="latex-formula" id="TexFormula3" src="https://tex.z-dn.net/?f=9%29%5C%3B%20%5C%3B%205%5Csqrt%7B18%7D-%5Csqrt2%2B0%2C2%5Csqrt8%3D5%5Ccdot%203%5Csqrt2-%5Csqrt2%2B0%2C2%5Ccdot%202%5Csqrt2%3D14%2C4%5Csqrt2%5C%5C%5C%5C%5C%5C10%29%5C%3B%20%5C%3B%20a%3D5%5C%3B%20%2C%5C%3B%20d%3D13%5C%3B%20%5C%3B%20%5Cto%20%5C%3B%20%5C%3B%20b%3D%5Csqrt%7B13%5E2-5%5E2%7D%3D%5Csqrt%7B144%7D%3D12%5C%5C%5C%5CS%3Dab%3D5%5Ccdot%2012%3D60%5C%5C%5C%5C%5C%5C11%29%5C%3B%20%5C%3B%20k%3E4%5C%3B%20%2C%5C%3B%20%5C%3B%20k%2B3-%5Csqrt%7B%284-k%29%5E2%7D%3Dk%2B3-%7C%5Cunderbrace%20%7B4-k%7D_%7B%3C0%7D%7C%3Dk%2B3-%28k-4%29%3D7" title="9)\; \; 5\sqrt{18}-\sqrt2+0,2\sqrt8=5\cdot 3\sqrt2-\sqrt2+0,2\cdot 2\sqrt2=14,4\sqrt2\\\\\\10)\; \; a=5\; ,\; d=13\; \; \to \; \; b=\sqrt{13^2-5^2}=\sqrt{144}=12\\\\S=ab=5\cdot 12=60\\\\\\11)\; \; k>4\; ,\; \; k+3-\sqrt{(4-k)^2}=k+3-|\underbrace {4-k}_{<0}|=k+3-(k-4)=7">

y(x)=3x+2

y(x)=0

0≠3x+2

-3x=2 (делим обе стороны на -3)

х= -2/3